solutions quadratic equation exercise # 1

QUADRATIC EQUATION

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H.O. 92, Rajeev Gandhi Nagar, Kota (Raj.) Mob. 97831-97831, 70732-22177, Ph. 0744-2423333 www.nucleuseducation.in

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SOLUTIONS – QUADRATIC EQUATION EXERCISE # 1

1. 6x2 – 7x + k = 0 have rational roots D = 49 – 24k, must be a perfect square k = 1 & 2 for k = 1, D = 25 for k = 2, D = 1 for no other value of k, D is perfect square

2. x2 – m(2x – 8) –15 = 0 � x2 – 2mx + 8m –15 = 0 D = 0 4m2 – 4 (8m –15) = 0 � m2 – 8m + 15 = 0 � (m – 5) (m – 3) = 0 m = 5 & 3

3. x2 – bx – c = 0

� D + E = ba

, DE =c

a�

then, D2 – DE + E2 = (D + E)2 – 3DE

= 2

2b 3c

aa�

=

2

2b 3ac

a�

4. ax2 + bx + c = 0 D < 0 � D = b2 – 4ac If b = 0 then a & c must have same sign. � b = 0, a > 0, c > 0

5. x2 + px + q = 0, let qy D� q

yD

� 2

q qp q 0y y

§ · § ·� � ¨ ¸ ¨ ¸

© ¹ © ¹ �

2

2q pq q 0

yy� �

� q2 + pqy + qy2 = 0 � y2 + py + q = 0

6. x2 + px + q = 0, D = p, E = q D + E = –p & DE = q � p + q = –p & pq = q � q (p –1) = 0 � 2p = –q & p =1 & q = 0

P = 0, Hence P = 1, 0

D E

QUADRATIC EQUATION

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7. ax2 + bx + c = 0

� � � �� a b a b

a b a b a b a bD D� �E E�D E

� E� D� E� D�

= � � � �

� �

2 22 2

2 2 2 2

ba b a ba ab ab b a ab b

D D �E � D�ED �D �E � E

DE� E� D� DE� D�E �

=

2

2 2

b 2c ba ba a ac ba ab ba a

§ ·� �§ · § ·� �¨ ¸¨ ¸ ¨ ¸¨ ¸© ¹ © ¹© ¹�§ · § ·� �¨ ¸ ¨ ¸

© ¹ © ¹

=

2 2

2

2 2

b 2ac baaa

ac b b

§ ·� ��¨ ¸

© ¹� �

= 2ac

ac a�u

= 2

a�

8. ax2 + x + b = 0

' D = 1– 4ab > 0 2x 4 abx 1 0� � D' = 16ab – 4 = –4(1– 4ab) = –4(+) = (–) . < 0 roots are imaginary.

9. x2 – 2x + 3 = 0 1 y1

D�

D ��

1 y1 y§ ·�

D ¨ ¸�© ¹

Then, 2

1 y 1 y2 3 01 y 1 y§ · § ·� �

� � ¨ ¸ ¨ ¸� �© ¹ © ¹

� (1+y)2 – 2(1+y) (1–y) +3(1–y)2 = 0 � 3y2 – 2y + 1 = 0 � 3x2 – 2x + 1 = 0, is required equation

10. x2 – 3x + 1 = 0

1 y

2

D � � D = 1 2y

y§ ·�¨ ¸© ¹

Then, 2

1 2y 1 2y3y 1 0y y

§ · § ·� �� ¨ ¸ ¨ ¸

© ¹ © ¹

� (1+2y)2 – 3y(1+2y) + y2 = 0 � y2 – y – 1 = 0 � x2 – x – 1 = 0, is required equation

D

E

QUADRATIC EQUATION

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11. 2x 5 0

x 5x 14�

!� �

� � ��

x 5 0x 7 x 2

�!

� � � � � � �x 7,2 5,� � � f

Least integral value = – 6

(A) D2 –7D + 6 = 0 (B) D2 + 3D – 4 = 0 � (D – 6) (D – 1) = 0 � (D – 1) (D + 4) = 0 � D = 6, D = 1 (C) D2 + 5D – 6 = 0 (D) D2 – 5D + 4 = 0 � (D – 1) (D + 6) = 0 � D2 – 4D – D + 4 = 0 � D = 1, D = –6 � D(D – 4) –1 (D – 4) = 0 � (D – 1) (D – 4) = 0

12. x2–3kx + 2log ke2e 1 0� , k > 0

� DE = 2log ke2e 1 7� � 2log kee 0 � k2 =4 � k = r 2 case – I k = + 2

� x2–6x + � �2log 2e2e 1 0� � � x2 – 6x + 8 – 1 = 0 x2 – 6x + 7 = 0 case – II k = –2 (Rejected) Check � D > 0 � 36 – 28 > 0 (Accepted) real roots

13. D + E = –p, DE = 3p4

10D�E � (D–E)2 = 10 � (D+E)2 – 4DE = 10

� p2 – 3p44

u = 10

� p2 – 3p – 10 = 0 � (p – 5) (p + 2) = 0 � P = {–2, 5}

14. D3 + E3 = –p & DE = q

� 2

' DD

E,

2' E

E D

2 –7 + – +

5 –

QUADRATIC EQUATION

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� D' + E' = 3 3D �EDE

= pq�

� D'E' = DE = q

Required equation x2 – pq

§ ·�¨ ¸© ¹

x + q = 0

� qx2 + px + q2 = 0

15. Equation 2x bx m 1

ax c m 1� �

� �

, D' + E' = 0

� x2 (m + 1) –bx(m + 1) –ax(m–1) + c(m–1) = 0 � x2 (m + 1) –x(a (m – 1) +b(m+1)) + c(m–1) = 0

� D' + E' = � � � �

� �b m 1 a m 1

0m 1

� � �

�

� bm + am = a – b

� m = a ba b�§ ·

¨ ¸�© ¹

16. y = x2 + ax + 25

this is upward parabola D = 0 � a2 – 100 = 0 � a = r10

17. a2x2 + bx + 1

D < 0 � b2 –4a2 < 0 � b2 < 4a2

18. y = ax2 + bx + c

a < 0

vertex b 0

2a�

!

� – b < 0 & f(0) > 0 � b > 0 c > 0

19. k(6x2 + 3) + rx + 2x2 – 1 = 0 & 6k(2x2 + 1) + px + 4x2 – 2 = 0 � x2(6k + 2) + rx + 3k – 1 = 0 & ��x2(12k + 4) + px + 6k – 2 = 0 Both roots common

� 6k 2 r 3k 112k 4 p 6k 2

� �

� � � r 1

p 2 � 2r – p = 0

QUADRATIC EQUATION

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20. x2 + 2 (a – 3) x + 9 = 0

f(6) < 0 � 36 + 2(a – 3)6 + 9 < 0 � 12a – 36 + 36 + 9 < 0

� 9a

12�

� �

3a4�

� ��3a ,

4�§ ·� �f¨ ¸

© ¹

EXERCISE # 2

1. x2 – px + q = 0

(i) 2 2

2 2§ · § ·D ED �E �E �D¨ ¸ ¨ ¸E D© ¹ © ¹

= � � � �2 2 2 2

2 2D �E E �D

D �EE D

= � �� 3 3 2 2D �E D �E

DE

= � �� 2 2D�E D �E �DE D�E D�E

DE

= � �� 22 2D�E D �E �DE D�E

DE

= � � � �� 2 2 4D�E D�E �DE D�E � DE

DE

=

� �� 2 2p p q p 4q

q

� �

(ii) (D – p)–4 + (E – p)–4

=

� � � �4 41 1

p p�

D� E� = � � � �

� � � �

4 4

4 4

p p

p p

E� � D�

D� E�

= � ��

4 4

4 4a p

E� D�E � D� D�E

D� � E� D�E

= � �

44

4D �E

DE =

� ��

22 2 2 2

4

2D �E � D E

DE

D E

D E

6 D� E

QUADRATIC EQUATION

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= � ��

� �

22 2

4

2 2D�E � DE � DE

DE =

� �22 2

4

p 2q 2q

q

� �

= 4 2 2 2

4p 4q 4p q 2q

q� � �

= 4 2 2

4p 2q 4p q

q§ ·� �¨ ¸© ¹

2. x2 + (2a–1)x + a2 + 2 = 0

Sum of roots D' + E' = – (2a–1) � D + 2D = 1 – 2a � 3D = 1 – 2a Product of roots D.2D = a2 + 2 � 2D2 = a2 + 2

� 2

21 2a2 a 23�§ · �¨ ¸

© ¹

� � �2

22 1 4a 4a

a 29

� � �

� 2 + 8a2 – 8a = 9a2 + 18 � a2 + 8a + 16 = 0 � (a + 4)2 = 0 a = – 4

3. x2 – 15 x4

+a = 0

Sum of roots D + D2 = 154

� 4D2 + 4D – 15 = 0 � (2D + 5) (2D – 3) = 0

� D = 5

2�

, D = 32

Put D = 5

2�

Put D = 32

� 25 75 a 04 8� � �

9 15 3 a 04 4 2� u �

� 125a8

� �

18 45a4 2

� �

u

� 27a8

D D2

D 2D

QUADRATIC EQUATION

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4. 5x2 – kx + 1 = 0

Sum of roots D + D – 1 = k5

� D = k 510�§ ·

¨ ¸© ¹

Product of roots D ( D – 1) =15

= D2 – D =15

= 5D2 – 5D –1 = 0

� 2k 5 k 55 5 1

10 10� �§ · § ·� ¨ ¸ ¨ ¸

© ¹ © ¹

� k 5 k 5 1110 10 5� �§ ·§ ·� ¨ ¸¨ ¸

© ¹© ¹

� k 5 k 5 110 10 5� �§ ·§ · ¨ ¸¨ ¸

© ¹© ¹

� k2 – 25 = 20 � k2 = 45 � k = r3 5

5. 5x2 + bx – 28 = 0

5D + 2E = 1 � 3D + 2(D + E) = 1

� b3 2 15�§ ·D � ¨ ¸

© ¹ �

5 2b15�§ ·D ¨ ¸

© ¹

Then, 25 2b 5 2b5 b 28 0

15 15� �§ · § ·� � ¨ ¸ ¨ ¸

© ¹ © ¹

� � � � �25 2b 3b 5 2b 28 45 0� � � � u � 2b2 + 7b – 247 = 0 b = – 13 Integer

6. y = x2 + ax + 25

D = 0 � a2 – 100 = 0 � a = r10

7. Vertex b D,2a 4a� �§ ·

¨ ¸© ¹

= � �24p 4 132p ,

2 4

§ ·� � u�¨ ¸¨ ¸© ¹

= (– p , 13–p2)

� � � �22 2p 13 p 5� � �

D E

D D–1

QUADRATIC EQUATION

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217

� p2 + (13–p2)2= 25 � p2 + 169 + p4 –26p2 = 25 � p4 – 25p2 + 144 = 0 � ( p2 – 9) ( p2 – 16) = 0 p = r 3 , p = r 4

8. ax2 + bx + c = 0 x1 + x2 =b

a�

(i) (ax1 + b)–2 + (ax2 + b)–2

� � � �2 2

1 1

1 1ax b ax b

��

= 2 22 2

1 2

1 1b ba x a xa a

�§ · § ·� �¨ ¸ ¨ ¸© ¹ © ¹

= � � � �2 22 2

1 1 2 2 1 2

1 1a x x x a x x x

��

= 2 2 22 1

1 1 1a x x

§ ·�¨ ¸

© ¹

= � �

� �

21 2 1 2

2 21 2

x x 2x x1a x x

§ ·� �¨ ¸¨ ¸© ¹

=

2

2

2 2

2

b 2caa1

a ca

§ ·§ ·�¨ ¸¨ ¸¨ ¸© ¹© ¹ =

2 2

2 2 21 a b 2aca c a

§ ·�u ¨ ¸

© ¹ = � �2

2 2

b 2ac

a c

�

(ii) (ax1 + b)–3 + (ax2 + b)–3

= � � � �3 3

1 2

1 1ax b ax b

��

= 3 33 3

1 2

1 1b ba x a xa a

�§ · § ·� �¨ ¸ ¨ ¸© ¹ © ¹

= � � � �3 3 3

2 1

1 1 1a x x

ª º« »�« »� �¬ ¼

= � �

3 31 2

3 31 2

x x1a x x

ª º�« »« »�¬ ¼

= � � � �

� �

31 2 1 2 1 2

3 31 2

x x 3x x x x1a x x

ª º� � �« »« »�¬ ¼

x1 x2

O

(–p, 13 – p2)

QUADRATIC EQUATION

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=

3

3

3 3

b c b31 a aaa c

a

ª º� �§ ·§ ·�« »¨ ¸¨ ¸© ¹© ¹« »

« »§ ·�« »¨ ¸© ¹¬ ¼

= � �2

3 3

b b 3ac

a c

�

9. k (x2 –x) + x + 5 = 0

kx2 + x (1–k) + 5 = 0 , k 1

5�§ ·D �E ¨ ¸

© ¹ ,

5k

DE

� 45

D E�

E D

� 2 2 4

5D �E

DE

� � �2 2 45

D�E � DE

DE

�

2k 1 104k k

5 5k

�§ · �¨ ¸© ¹

� � �2

2

k 1 10k 45 5kk

� �

u

� 2k 2k 1 10k 4

5k 5� � �

� k2 – 12k + 1 = 4k

� k2 – 16k + 1 = 0

1 2

2 1

k kk k

� = 2 2

21

1 2

k kk k�

= � �21 2 1 2

1 2

k k 2k kk k

� � =

� �216 2 11� u

= 254

10. ax2 + bx + c = 0

Ax2 + Bx + C = 0 D + G

E + G

D E

k1 k2

D E

QUADRATIC EQUATION

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219

Prove that 2

2b 4ac

a� =

2b c4a a

§ · § ·�¨ ¸ ¨ ¸© ¹ © ¹

= (D + E)2 – 4DE = (D – E)2

2

2B 4AC

A� =

2

2B 4ACA A

§ · �¨ ¸© ¹

= 2B C4

A A§ · § ·�¨ ¸ ¨ ¸© ¹ © ¹

= (D + G + E + G )2 – 4 ( D + G ) (E + G ) = ((D + G ) – (E + G ))2 = (D – E)2 Hence Proved

11. (a) ax2 + bx + c = 0 (i) f (D, E) = D2 – E f(E, D) = E2 –D � f (D, E) z f (E, D) (ii) if (D, E) = D2E +DE2 , f(E, D) = E2D + ED2 f (D, E) = f (E, D)

(iii) f(D, E) = ln DE

= –f (E, D) z f (E, D)

(iv) f(D, E) = cos (D – E) & f(E, D) = cos (E – D) f (D, E) = f (E, D)

(b) x2 – px + q = 0

D + E = p, DE = q J = (D2 – E2) (D3 – E3) = (D + E) (D – E) (D –E) (D2 +E2 + DE) = (D – E)2 (D + E) (D2 +E2 + DE) = ((D + E)2 – 4DE) (D + E) ((D +E)2 – DE) = (p2 –4q) ( p ) (p2 – q) = (p4 –4p2q – p2q + 4q2) p = (p4 –5p2q + 4q2) p = (p5 –5p3q + 4q2p)

G = D3E2 + D2E3 = D2E2 (D + E) = q2 (p) = pq2

J + G = p5 – 5p3q + 4q2p + pq2 = p5 – 5p3q + 5pq2 J G = pq2 (p5 – 5p3q + 4pq2) = p2q2 (p4–5p2q + 4q2) x2 – (p5 – 5p3q + 5pq2) x + p2q2 (p4 – 5p2q + 4q2) = 0 x2 –p (p4 – 5p2q + 5q2) x + p2q2 (p4 – 5p2q + 4q2) = 0

D E

QUADRATIC EQUATION

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12. (i) 2

2x 2x 3 0

x 1� �

��

, x2 + 1 is always positive

� x2 + 3x – x – 3 < 0 � x (x + 3) –1 (x + 3) < 0 � (x +3) (x – 1) < 0 x � (–3, 1)

(ii) � �� 2x 1 x 20

1 x� �

��

� � ��

2x 1 x 20

x 1� �

!�

� � � � � �x , 2 2, 1 1,� �f � � � � � f (iii) x4 – 2x2 – 63 ≤ 0 � x4 – 9x2 + 7x2 – 63 ≤ 0 � x2 (x2 – 9) + 7 (x2 – 9) ≤ 0 � (x2 – 9) (x2 + 7) ≤ 0 � (x + 3) (x – 3) (x2 + 7) ≤ 0 x � [–3, 3]

(iv) � �� 2

x 11

x 1

��

�

� � ��

2

2

x 1 x 10

x 1

� � ��

�

� � �

2

2x 1 x 1 2x 0

x 1� � � �

��

� � �

2

2x 3x 0x 1

� ��

�

� � �

2

2x 3x 0x 1�

!�

� � �� 2

x x 30

x 1

�!

� � � � �x ,0 3,� �f � f

(v) 2

2x 7x 12 02x 4x 5

� �!

� �, 2x2 + 4x + 5 is always positive

� (x – 3) (x – 4) > 0 � � � �x ,3 4,� �f � f

(vi) 2

2x 6x 7 2

x 1� �

d�

, x2 + 1 is always positive

1 –3 – + +

1 –2 – + + +

–1

3 –3 – + +

1 0 – + +

3 –

4 3 – + +

QUADRATIC EQUATION

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221

� 2 2

2x 6x 7 2x 2 0

x 1� � � �

d�

� –x2 + 6x – 9 ≤ 0 � x2 – 6x + 9 t 0 � (x – 3)2 t 0 always true. � �x ,� �f f

(vii) 4 2

2x x 1 0x 4x 5

� ��

� �, x4 + x2 + 1 is always positive

� � ��

0x 5 x 1

��

� �

� �x 1,5� �

(ix) 1 3

x 2 x 3�

� �

� 3 0

x 2 x 3�

� ��

� � ��

2x 9 0x 2 x 3

�!

� �

� �9x , 2 3,2

§ ·� � � � f¨ ¸© ¹

(viii) x 7 3x 1 0x 5 2� �

� t�

� � �

22x 14 3x x 15x 5 02 x 5

� � � � �t

�

� � �

23x 12x 9 02 x 5� �

t�

� � ��

� �x 3 x 1

02 x 5� �

t�

> @ � �x 1,3 5,� � f

(x) 14x 9x 30 0x 1 x 4

��

� �

� � � � ��

� �� 14x x 4 3 3x 10 x 1

0x 1 x 4

� � � ��

� �

5 1 – + + –

3

3 92�

– + + – –2

5 –1 – + +

QUADRATIC EQUATION

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222


222

� � �

� ��

2 214x 56x 3 3x 7x 100

x 1 x 4

� � � ��

� �

� � ��

25x 35x 30 0x 1 x 4� �

��

� � ��

2x 7x 6 0x 1 x 4

� ��

� �

� � �� x 6 x 1

0x 1 x 4� �

��

� � � �x 1,1 4,6� � �

(xi) 2

2x 5x 12 3x 4x 5� �

!� �

, x2 – 4x + 15 is always positive

� 2 2

2x 5x 12 3x 12x 15 0

x 4x 15� � � � �

!� �

� (x – 3) (2x – 1) < 0

1x ,32

§ ·�¨ ¸© ¹

(xii) 2

2x 2 2x 1�

� ��

� � �� 2 2x 2 2x 2 0x 1 x 1� � �

��

� � ��

23x 0x 1 x 1

��

� � � �x 1,0 0,1� � �

(xiii) � ��

� �� 32

2

2 x x 30

x 1 x 3x 4

� �t

� � �

� � ��

� ��

32x 2 x 30

x 1 x 4 x 1

� �d

� � �

� � ��

� � � �

3

2

x 2 x 2 x 30

x 1 x 4

� � �d

� � � � > �x 2, 1 1, 2 3,4ª º� � � � � �¬ ¼

(xiv) 25 4x 4

3x x 4�

��

� � �2

2

5 4x 4 3x x 40

3x x 4

� � � ��

� �

� 24x 7 0

x(3x 4) 1(3x 4)�

!� � �

2– 2 – – + +

–1 – +

3 4

1 –1 – – + +

0

3 – + +

12

4 –1 + – – +

1 +

6

QUADRATIC EQUATION

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223


223

� 2 7x

4 0(3x 4)(x 1)

�!

� �

�

7 7x x2 2

0(3x 4)(x 1)

§ ·§ ·� �¨ ¸¨ ¸

© ¹© ¹ !� �

7 7 4x , 1, ,2 2 3

§ · § · § ·� �f � � � � f¨ ¸ ¨ ¸ ¨ ¸¨ ¸ ¨ ¸ © ¹© ¹ © ¹

(xv) � �� 2

2

x 2 x 2x 10

4 3x x

� � �t

� �

� � ��

2

2

x 2 x 10

x 3x 4� �

t� � �

� � ��

2x 2 x 10

x 4 x 1� �

d� �

� @ � �x , 2 1,4� �f � � �

(xvi) 4 3 2

2x 3x 2x 0

x x 30� �

!� �

� � ��

� ��

2x x 2 x 10

x 6 x 5� �

!� �

� � � � � �x , 5 1,2 6,� �f � � � f

(xvii) 22x 1

x 2x 9d

��

� 22x 1 0

x 2x 9� d

��

� � �

� �� 22x x 2 x 9

0x 3 x 3 x 2

� � �d

� � �

� � ��

2x 4x 9 0x 3 x 3 x 2

� �d

� � �, x2 + 4x + 9 is always positive

� � � �x , 3 2,3� �f � � �

(xviii) 1 1 1

x 2 x 1 x� !

� �

� 1 1 1 0

x 2 x 1 x� � !

� �

3 –3 – + + –

–2

0 + – – –

1 2 6 +

–5 +

–2 – + – –

–1 1 4 +

72

�

+ – – + –1 7

2 4

3

+

QUADRATIC EQUATION

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224


224

� � � � � � ��

� �� x x 1 x 2 x x 2 x 1

0x 2 x 1 x

� � � � � �!

� �

� � �

� ��

2 2 2x x x 2x x 2x x 20

x x 1 x 2

� � � � � � �!

� �

� � ��

2x 2 0x x 1 x 2

�!

� �

� � ��

� �� x 2 x 2

0x x 1 x 2

� �!

� � � � � � � �x 2,0 1, 2 2,� � � � f

(xix) � ��

20 10 1 0x 3 x 4 x 4

� � !� � �

� � � � ��

20 10 x 3 x 3 x 40

x 3 x 4� � � � �

!� �

� � �

� ��

220 10x 30 x 7x 120

x 3 x 4

� � � � �!

� �

� � ��

2x 3x 2 0x 3 x 4

� �!

� �

� � �� x 2 x 1

0x 3 x 4� �

!� �

� � � � � �x , 2 1,3 4,� �f � � � � f

(xx) � �� x 2 x 4 x 7

1x 2 x 4 x 7� � �

!� � �

� � ��

� �� x 2 x 4 x 7 x 2 x 4 x 7

0x 2 x 4 x 7

� � � � � � �!

� � �

� � ��

� ��

2 2x 6x 8 x 7 x 6x 8 x 70

x 2 x 4 x 7

� � � � � � �!

� � �

� � ��

26x 7x 56 0x 2 x 4 x 7

� ��

� � �, 6x2 + 7x + 56 is always positive

� � � � �x , 7 4, 2� �f � � � �

(xxii) (x2–2x) (2x–2) – � �

2

9 2x 20

x 2x�

d�

–7 – + + –

–4 –2

-1 – + +

–2 –

3 4 +

1 – 2 – + + –

0 –

2 2 +

QUADRATIC EQUATION

225

225


225

� � � � � � �

� �

22x 2x 2x 2 9 2x 20

x x 2

� � � �d

�

� � � � ��

� �

222 x 1 x 2x 90

x x 2

� � �d

�

� � ��

� �

2 22 x 1 x 2x 3 x 2x 30

x x 2

� � � � �d

�

� � ��

� �2 x 1 x 3 x 1

0x x 2

� � �d

� � @ � @ � @x , 1 0,1 2,3� �f � � �

13. (k–12)x2 + 2(k–12)x + 2 = 0

' D < 0 For no real root � 4(k–12)2 – 8(k–12) < 0 � (k –12) (k–12–2) < 0 � (k–12) (k–14) < 0

k = 13 only possible integral value.

14. x2 – (k–3)x – k + 6 > 0

D < 0 � (k–3)2 – 4(–k+ 6) < 0 � k2 – 6k + 9 + 4k – 24 < 0 � k2 – 2k –15 < 0 � (k – 5) (k + 3) < 0 k � (–3, 5)

15. (P–3)x2 – 2px + 5p = 0 real and positive roots.

Case–I Case–II Case–III p–3 > 0 p – 3 < 0 p =3 p > 3 p < 3

case – I D t 0, f(0) > 0, b 0

2a�

!

For D t��

14 12 + – +

–1 – + + –

1 2 0 – +

3

5 –3 + – +

QUADRATIC EQUATION

226

226


226

� 4p2 – 4(p–3) 5p t 0

� p2 – (p–3) 5p t 0 ………………(A)

� p2 – 5p2 + 15p t 0 � 4p2 – 15p ≤ 0 � p(4p –15) ≤ 0 for f(0) > 0 p > 0…………..(B)

For b 0

2a�

!

� �2p 0

2 p 3!

�…………(C)

From equation (A), (B) & (C)

15p 3,4

§ º�¨ »© ¼

Case –2 P < 3

D t O � 15p p 04

§ ·� d¨ ¸© ¹

f(0) < 0, P < 0 P � I Case – 3 P = 3 (P –3)x2 –2px + 5p = 0 –6x + 15 = 0

x = 52

taking union of all three cases.

15P 3,4

ª º�« »¬ ¼

16. (a+4)x2 – 2ax + 2a – 6 < 0 for all x � R � (a+4) < 0 � a < – 4 ……….(1) D < 0 � 4a2 – 4(a + 4) (2a – 6) < 0 � a2 – (2a2 + 2a–24) < 0 � –a2 – 2a + 24 < 0 � a2 + 2a – 24 > 0 � a2 +6a – 4a – 24 > 0 (a + 6) (a – 4) > 0…………….(2)

from equation (1) & (2), a � (–f, –6)

–6 4 + – +

154

0 + – +

3 0 154

QUADRATIC EQUATION

227

227


227

17. y = 2ax + 1 and y = (a – 6)x2 –2 no solution. � 2ax + 1 = (a – 6)x2 –2 � (a – 6)x2 – 2ax – 3 = 0 , D < 0 � 4a2 + 4 × 3 (a – 6) < 0 � a2 + 3a – 18 < 0 � (a – 3) (a + 6) < 0

a � (–6, +3)

18. f(x) = 2

2ax 2(a 1) x 9a 4 0

x 8x 32� � � �

��

, x2 – 8x + 32 is always positive

� ax2 + 2(a+1) x + 9a + 4 < 0 a < 0 …………………..(1) D < 0 � 4 (a+1)2 – 4a (9a + 4) < 0 � a2 + 2a + 1 – 9a2 – 4a < 0 � –8a2 – 2a + 1 < 0 � 8a2 + 2a –1 > 0 � (4a – 1) (2a + 1) > 0 ………….(2)

From equation (1) & (2)

a � 1,2�§ ·�f¨ ¸

© ¹

19. x2 + 3x – k = 0

x2 + 3x – 10 = 0

� |D – E| = 2 |c – d| � (a–b)2 = 4(c–d)2

� a2 + b2 – 2ab = 4 (c2 + d2 –2cd) � (a + b)2–4ab = 4 ((c+d)2 – 4cd) � 9 + 4k = 4 ( 9 + 40)

k = 187

4, then m + n = 187 + 4 = 191

20. Let D is common root

x2 –11x + m = 0 � D2 – 11D + m = 0 � x2 – 14x + 2m = 0 � D2 – 14D + 2m = 0

c

d

a

b

12�

0

14

–6 3 + – +

QUADRATIC EQUATION

228

228


228

� 2 1

11 m 1 m 1 1114 2m 1 2m 1 14

D �D

� ��

� 2 1

22m 14m m 3D �D

� � �

� D =m3

& D2 =

8m3

� 2m 8m

9 3

� m2 – 24m = 0 � m( m – 24) = 0 m = 0 or m =24

21. x2 + bx + ca = 0 D2 + bD + ca = 0

x2 + cx + ab = 0 D2 + cD + ab = 0

� 2 1

b ca 1 ca 1 bc ab 1 ab 1 c

D �D

� � � � �

2

2 21

a b c c bab acD �D

� ��

………….(1)

� D = � ��

a c bc b��

� D = a common root ……….. (2)

from equation (1) D2 = � ��

a b c b cc b� ��

� –a2 = � ��

a b c b cb c� ��

(Put D = a)

� a = – (b + c) Sum of roots D + E = –b E = – b –a E = a + c –a E = c Sum of roots D + J = –c J = – c – D = – c – a = – c + b + c � J = b x2 – (E + J)x + EJ = 0 � x2 – (b + c)x + bc = 0 x2 + ax + bc = 0 Hence proved

D

J

D

E

QUADRATIC EQUATION

229

229


229

22. f(x) = x2 – 4ax + 5a2 – 6a (a) D < 0 � 16a2 – 4(5a2 – 6a ) < 0 � 4a2 – 5a2 + 6a < 0 � –a2 + 6a < 0 � a2 – 6a > 0 � a (a – 6) > 0 smallest positive integer = 7 (b) x2 – 4ax + 5a2 – 6a

� x1 , x2 = � �2 24a 16a 4 5a 6a

2

r � �

= 2 22a 4a 5a 6ar � �

= 22a 6a ar �

& (x1 – x2) = � �22a 6a a� � – � �22a 6a a� �

= 22 6a a� = � �22 a 6a 9 9� � � �

� (x1 – x2) = � �22 9 a 3� � � (x1 – x2)max = 2 × 3 = 6 for a = 3

(c) D 8

4a�

�

� � �2 216a 4 5a 6a

84

� �

� 4a2 – 5a2 + 6a = 8 � a2 – 6a + 8 = 0 � (a – 4) (a – 2) = 0 � a = 4 & a = 2

23. 2

22x 2x 3 p

x x 1� �

d� �

� 2 2

22x 2x 3 px px p 0

x x 1� � � � �

d� �

, x2 + x + 1 is always positive

D < 0

� x2 (2–p) + x(2–p) + (3–p) ≤ 0 � 2 – p < 0 p >2 & D ≤ 0 � (2 – p)2 – 4(2–p) (3–p) ≤ 0 � (2 – p) (2–p–4 (3–p)) ≤ 0 � (2–P) (2–P–12+4P) ≤ 0 � (2–P) (3P–10) ≤ 0

6 0 + – +

–8

QUADRATIC EQUATION

230

230


230

� (P–2) (3P–10) t 0

� P � 10 ,3

ª ·f¸«¬ ¹

smallest integral = 4

24. x3 – 3x2 + 1 = 0

� y2

D

D�,

2yy 1

§ ·D ¨ ¸�© ¹

x3 – 3x2 + 1 = 0

� 3 2

2y 2y3 1 0y 1 y 1

§ · § ·� � ¨ ¸ ¨ ¸� �© ¹ © ¹

� 8y3 – 3 × 4y2 (y–1) + (y –1)3 = 0 � –3y3 + 9y2 + 3y – 1 = 0 � 3y3 – 9y2 – 3y + 1 = 0 � 3x2 – 9x2 –3x + 1 = 0 (D – 2) (E – 2) (J – 2) = (DE –2 (D +E) + 4) (J – 2) = DEJ – 2J (D+E) + 4J – 2DE + 4 (D+ E) – 8 = –1 –2 (DJ + JE + DE) + 4(D +E +J) – 8 = –1 –2 (0) + 4(3) – 8 = 3

25. � 2x

x 5x 9� � = y

� x = yx2 – 5xy + 9y � yx2 – x (5y + 1) + 9y = 0 D t 0 � (5y + 1)2 – 4 × y × 9y t 0 � 25y2 + 1 + 10y – 36y2 t 0 � –11y2 + 10y + 1 t 0 � 11y2 – 10y – 1 ≤ 0 (11y + 1) (y – 1) ≤ 0

26. � 2x 2

2x 3x 6��

= y

� 2yx2 + x(3y–1) + 6y –2 = 0 D t 0 � 9y2 + 1 – 6y – 4 × 2y × (6y – 2) t 0 � 39y2 – 10y – 1 ≤ 0

D

E

J

2 + – +

103

1 + – +

111�

+ – + 1

13� 1

3�

QUADRATIC EQUATION

231

231


231

� (13y +1) (3y –1) ≤ 0

Greatest value = 13

27. m = ? y2 + 2xy + 2x + 3y – 3 be capable of resolution into two rational factors. D t 0 y2 + y(2x + m) + 2x –3 = 0 � (2x +m)2 – 4 (2x–3) t 0 � 4x2 + 4x (m–2) + m2 + 12 t 0 � 16(m–2)2 – 4 × 4 (m2 + 12) ≤ 0 ('D d��) � m2 + 4 – 4m – m2 – 12 ≤ 0 m t – 2

28. 9x2 + 2xy + y2 – 92x – 20y + 244 = 0 � y2 + 2y (x–10) + (9x2 – 92x + 244) = 0 D t 0 � 4(x–10)2 – 4 (9x2 – 92x + 244) t 0 � x2 +100 – 20x –9x2 + 92x – 244 t 0 � –8x2 + 72x – 144 t 0 � (x2 – 9x + 18) ≤ 0 � (x–3) (x–6) ≤ 0 3 ≤ x ≤ 6 Proved similarly � 9x2 + 2x (y – 46) + (y2 – 20y +244) = 0 D t 0 � 4(y – 46)2 – 4 × 9 (y2 –20y + 244) t 0 � y2 +462 – 92y – 9y2 + 180y – 9 × 244 t 0 � –8y2 + 88y + 462 – 9 × 244 t 0 � 8y2 – 88y + 80 ≤ 0 � y2 –11y + 10 ≤ 0 � (y–10) (y–1) ≤ 0

1 ≤ y ≤ 10 Proved

29. (a2 – 6a + 5)x2 – 2a 2a� x + (6a –a2– 8) = 0

� a2 – 6a + 5 > 0 & a2 + 2a t 0 � (a–5) (a–1) > 0 & a (a + 2) t 0 Case (1) Case (II) � a < 1 & a > 5 ………..(1) 1 < a < 5 � f (0) < 0 f (0) > 0 � 6a – a2 – 8 < 0 (a – 4 ) (a –2) < 0 (a – 4 ) (a –2) > 0……….(2) � �a 2,4� ……(4) From equation (1) & (2)

+ – + 1 10

+ – + 3 6

o

QUADRATIC EQUATION

232

232


232

� � @ > � � �a , 2 0,1 5,� �f � � � f ………(3) from equation (3) & (4) � � @ > � � � � �a , 2 0,1 2,4 5,� �f � � � � f

30. f(0) > 0 f(3) > 0 D t 0

always true 9–3a + 2 > 0 a2 – 8 t 0

a < 113

……(1)

� �� a 2 2 a 2 2 0� � t ……………..(2)

B0 3

2A�

� �

a0 32

� �

0 < a < 6 ……………..(3)

From equation (1), (2) & (3), Common solution – 11a 2 2,3

ª ·� ¸«¬ ¹

EXERCISE # 3 (JM)

1. x2 – x(a – 2)– a – 1 = 0 � D2 + E2 = (D + E)2 – 2DE�� = (a – 2)2 + 2(a + 1) � y = a2 – 2a + 6 � for maxima or minima. y' = 2a – 2 = 0 ��a = 1

2. x2 – bx + c = 0

� 2b b 4c,

2r �

D E for two consecutive integers, b2 – 4c must be perfect square.

only possibility = 1

3. x2 –2kx + k2 + k – 5 = 0 D t 0 � 4k2 – 4 (k2 + k – 5) t 0 � k2 – k2 – k + 5 t 0 � k d 5………………(1) f(5) > 0 � 25 – 10k + k2 + k – 5 > 0 � k2 – 9k + 20 > 20 � (k – 4) (k – 5) > 0 …………(2)

3 0

5� b3

� �

QUADRATIC EQUATION

233

233


233

� b 5

2a�

� � 2k 52�

�

k < 5 …………(3)

after taking intersection of conditions (1), (2) & (3), k � (–f, 4)

4. x2 – 2mx + m2 –1 = 0

f(–2) > 0 f(4) > 0 � 4 + 4m + m2 –1 > 0 16 – 8m + m2 – 1 > 0 D t�0 � m2 + 4m + 3 > 0 m2 – 8m + 15 > 0 4m2 – 4(m2 – 1) t 0 � (m + 3) (m +1) > 0 (m –5) (m – 3) > 0 m2 – m2 + 1 t 0 � 1 t 0

…………..(1) …………….(2) m � R ……….(3)

�b2 4

2a�

� � �

��2m2 42

� � � �� m < 8

� – 2 < m < 4………(4)� From equation (1), (2) & (3) , (4) we get, –1 < m < 3

5. x2 + px +q = 0

tan30º + tan15º = – p tan30º tan15º = q � 2 + q – p = 2 + tan30o tan15o + tan30o + tan15º � � � � � 2 + tan30o ( 1+ tan15o )+ tan15º

� � � � � � �12 1 2 3 2 33

� � � � �

� � � � 3 34 33

��

� � � � 4 3 1 3� � � � � � � 3

6. Let 2

23x 9x 17y3x 9x 7

� �

� �

� 2

23x 9x 7y3x 9x 7

� �

� � + 2

103x 9x 7� �

1 + 210

3x 9x 7� �

for y max., 3x2 + 9x + 7 must be minimum

tan30o

tan15º

4�–2�

5 + – +

3 –1 –3 + – +

QUADRATIC EQUATION

234

234


234

For minimum value, � �2b 4acD

4a 4a

� ��

� � � � � �� 81 4 3 7

12� � u u

� � � � � �� 81 84

12� �

� � � � � �14

� � �max

10y 11

4 � = 41

7. x2 + ax + 1 = 0

� |D – E | < 5 � D2 + E2 – 2DE < 5 � (D + E)2 – 4DE < 5 � a2 – 4 < 5 � a2 – 9 < 0 � (a + 3) (a –3) < 0

a � (–3, 3)

8. x2 – 6x + a = 0 x2 – cx + 6 = 0

� D + E = 6 D + J = c � DE = a DJ = 6

a 46 3

DE

DJ (given)

� a = 8 D + E = 6

� D + 8D

= 6

� D2 – 6D + 8 = 0 � D = 4, 2

43

E

J � 3E = 4J

D J D E

–3 + – +

3

QUADRATIC EQUATION

235

235


235

��4232

½°D °

E ¾°°J ¿

α =2β =4γ =3 integer

½°¾°¿

then common root is 2�

9. bx2 + cx + a = 0

for imaginary roots D < 0 � c2 – 4ab < 0 � c2 < 4ab � –c2 > – 4ab Expression 3b2x2 + 6bcx + 2c2 represents upword parabola

� �2 2 2 2

2

36b c 4 3b 2cD4a 4 3b

� � u u�

u� �2 2

22

12b cc

12b

� � , greater than –4ab

10. P(x) = 0 � f(x) = g(x) � ax2 + bx + c = a1x2 + b1x + C1 � (a – a1) x2 + (b – b1) x + (c – c1) = 0. It has only one solution x = –1 � b – b1 = a – a1 + c – c1 …(1)

Vertex (–1, 0) � 1

1

b b 12(a a )

� �

� � b – b1 = 2(a – a1) …(2)

� f(–2) – g(–2) = 2 � 4a – 2b + c – 4a1 + 2b1 – c1 = 2 � 4(a – a1) – 2(b – b1) + (c – c1) = 2 …(3)

by (1), (2) and (3) (a – a1) = (c – c1) = 12

(b – b1) = 2

Now P(2) = f(2) – g(2) = 4 (a – a1) + 2 (b – b1) + (c – c1) = 8 + 8 + 2 = 18

11. Sachin reads sum of roots correctly = 7 Rahul reads product of roots correctly = 6 Hence required equation is x2 – 7x + 6 = 0 � x = 6, 1

12. esinx – e–sinx – 4 = 0 y = esinx can’t be negative

� � �2sin x sin xe 4e 1 0� � esinx = 2 + 5

� y2 – 4y – 1 = 0 (sinx) cannot be greater than unity, so no solution.

� 4 16 4y 2 52

r � r

–c2�

QUADRATIC EQUATION

236

236


236

13. x2 + 2x + 3 = 0 ax2 + bx +c = 0 D < 0 so imaginary roots, so both roots are common

� 1 2 3a b c

� a : b : c = 1 : 2 : 3

14. x2 – 6x – 2 = 0 an = Dn – En n t 1

� x8 (x2 – 6x – 2) = 0 � x10 – 6x9 – 2x8 = 0 � D10 – 6D9 – 2D8 = 0 ……….(1) E10 – 6E9 – 2E8 = 0………..(2)

� (D10 – E10) + 6(E9 – D9) + (E8 – D8) = 0

� � � � �

� �10 10 8 8

9 9

23

2

D �E � D �E

D �E

15. (x2 – 5x + 5)x2 + 4x – 60 = 1

case –(1) � x2 – 5x + 5 = 1 � x2 – 5x + 4 = 0 � (x – 4) (x – 1) = 0 � x = 4, 1 case – (2) x2 – 5x + 5 = –1 � x2 – 5x + 6 = 0 � (x – 3) (x – 2) = 0 � x = 3, 2 For x = 3 (x2 + 4x – 60) = 9 + 12 – 60 = –39 odd power (Rejected) For x = 2 (x2 + 4x – 60) = 4 + 8 – 60 = 48 even power (Accepted) case – (3) x2 + 4x – 60 = 0 � x2 + 10x – 6x – 60 = 0 � x(x +10) – 6(x +10) = 0 � (x –6) (x +10) = 0 � x = 6 x = –10 sum of all real values of x = 4 + 1 + 2 + 6 – 10 13 – 10 3

D

E

④

④

⑧

④

⑧

④

④"

④

④

④

⑥

④

⑧

④

⑧

③

④

QUADRATIC EQUATION

237

237


237

EXERCISE # 4 (JA)

1. � 2

21 2x 5x3x 2x 1� �

� � = y

� 1–2x�+ 5x2 = 3yx2–2xy–y � x2 (3y–5) – 2x(y–1) – (y+1) = 0 D t 0 � 4(y–1)2 + 4(3y–5)(y+1) t 0 � y2–2y+1 + 3y2+3y–5y– 5 t 0 � 4y2–4y–4 t 0 � y2–y–1 t 0

� 1 5 1 5y y2 2

§ ·§ ·§ · § ·� �� ¨ ¸¨ ¸¨ ¸ ¨ ¸¨ ¸ ¨ ¸¨ ¸¨ ¸© ¹ © ¹© ¹© ¹

t 0

�

� y d 1 52

§ ·��¨ ¸¨ ¸© ¹

or y t 1 52

§ ·�¨ ¸¨ ¸© ¹

� sint d� 1 54

§ ·�¨ ¸¨ ¸© ¹

or sint t� 1 54

§ ·�¨ ¸¨ ¸© ¹

� Hence range of t is 3, ,2 10 10 2�S �S S Sª º ª º�« » « »¬ ¼ ¬ ¼

2. (a) a, b, c be the sides of a triangle, No two of them are equal

� x2 + 2(a + b + c)x + 3O(ab + bc + ca) = 0 D t 0 � 4(a + b + c)2 – 12O (ab + bc + ca) t 0 � (a + b + c)2 – 3O (ab + bc + ca) t 0

� O d � ��

2a b c3 ab bc ca

� ��

� O d��

2 2 2a b c3 ab bc ca

� ��

+ 23

� |a –b| < c � a2 + b2 – 2ab < c2 ………….(1) � |b –c| < a � b2 + c2 – 2bc < a2 ………….(2) � |c –a| < b � c2 + a2 – 2ca < b2 ………….(3) from equation (1), (2) & (3)

� � �

2 2 2a b c 2ab bc ca� �

��

1 52

§ ·�¨ ¸¨ ¸© ¹

1 52

§ ·�¨ ¸¨ ¸© ¹

+ + –

QUADRATIC EQUATION

238

238


238

� 2 23 3

O � �

� 43

O �

(b) x2 – 10cx – 11d = 0

x2 – 10ax – 11b = 0

Sum of roots, a + b = 10c & c + d = 10a � a + b + c +d = 10 (a + c) � b + d = 9(a + c ) ……………(1) Product of roots, ab = 11d & cd = –11b � abcd = 121 bd � ac = 121 …………….(2) Now, a2 – 10ac – 11d = 0 c2 – 10ac – 11b = 0 � (a2 + c2) – 20ac – 11(b + d) = 0 � (a2 + c2) – 20ac – 11× 9(a + c) = 0 � (a + c)2 – 22ac – 99 (a + c) = 0 � (a + c)2 – 99(a + c) –22 × 121 = 0 � (a + c) = 121 or –22 (rejected) a + c = 121 b + d = 9 × 121 a + b + c + d = 1210

3. (a) x2 – px + r = 0

x2 – qx + r = 0

2 q2D� E & D + E = p & DE = r

� D + 4E = 2q � 3E = 2q – p ��D = p – E

� E = 2q p3�§ ·

¨ ¸© ¹

� D = 2q pp3�§ ·�¨ ¸

© ¹

�

4p 2q3�§ ·D ¨ ¸

© ¹

DE = r

� � �� 2 2p q 2q p

r9

� �

� � �� 2r 2p q 2q p9

� �

D/2

2E

D

E

c

d

a

b

QUADRATIC EQUATION

239

239


239

(b) � �2

2x 6x 5f x yx 5x 6

� �

� �

� �2

2x 6x 5f xx 5x 6

� �

� �

� � � �

� �2

2

1 x 5x 6 x 1

x 5x 6

� � � �

� �

� = � ��

x 11

x 2 x 3�

��

(A) if – 1 < x < 1 � 0 < f(x) < 1 � f(x) > 0 (A) ��o p ,r ,s � f(x) < 1 (B) if 1 < x < 2 � f(x) < 0 � f(x) < 1 (B) ��o q, s (C) if 3 < x < 5 � f(x) < 0 � f(x) < 1 (C) ��o q, s (D) if x > 5 � 0 < f(x) < 1 � f(x) > 0 (D) ��o p, r ,s � f(x) < 1

4. x2 + 2px + q = 0

ax2 + 2bx + c = 0 & E2 � {–1, 0, 1}

suppose roots are imaginary, then E D and 1 D

E

� 1E

E not possible . [' E2 � {–1, 0, 1}]

roots are equal � D1 t 0 D2 t�0 � 4p2 – 4q t 0 4b2 – 4ac t 0 � (p2 – q) (b2 – ac) t 0 statement (1) is true.

D

1/E

D

E

1 2 3 4

y = 1

QUADRATIC EQUATION

240

240


240

� 1 2ba�

D� E

& ca

D

E

� D + E = –2p DE = q statement (2) ' one root is common

� 2

2pc 2bqD�

= aq cD�

= 12b 2ap�

for defining of D

b z ap, aq z c � statement (1), (2) both true. But (2) is not a correct explanation for statement (1).

5. x2 –8kx +16(k2 –k +1) = 0 f(4) t�0 � 16 – 32k + 6(k2 – k + 1) t 0 � 1 – 2k + ( k2 – k + 1) t 0 � k2 –3k + 2 t 0 � (k–2) (k–1) t 0……………(1) D t 0 � 64k2 – 64(k2 – k +1) t 0 � k t 1………………(2)

b 4

2a�

!

� 8k 42!

� k > 1 …………(3) from equation (1), (2) & (3) � > �k 2,� f Hence smallest value = 2

6. p z 0, p3 z q and p3 z – q � D + E = – p, D3 + E3 = q New roots

sum = D E�

E D =

2 2D �EDE

= � �2

2D�E

�DE

=

2

33p q 2p q

��

=

3 3

33p 2p 2q

p q� �

�

4�

QUADRATIC EQUATION

241

241


241

=

3

3p 2qp q��

(D + E)3 = D3 + E3 + 3DE (D + E)

� 3p q3p

� � DE

� ��DE� �

3p q3p

§ ·�¨ ¸© ¹

New required equation

� 3

23

p 2qx x 1 0p q

§ ·�� ¨ ¸�© ¹

� 3 2 3 3(p q) x (p 2q) x (p q) 0� � � � �

7. x2 – 6x – 2 = 0 D2 – 6D – 2 = 0 D10 – 6D9 – 2D8 = 0 …………(1) E10 – 6E9 – 2E8 = 0 …………(2) � (D10–E10) + 6 (E9 – D9) + 2 (E8 –D8) = 0

� � � � �

� �10 10 8 8

9 9

23

2

D �E � D �E

D �E

8. x2 + bx – 1 = 0

x2 + x + b = 0 let D is common root. � D2 + bD – 1 = 0 � D2 + D + b = 0

2

b 11 b

D�

= 1 11 b

�D� =

11 b1 1

��2

2b 1D�

= b 1�D�

= 1

1 b�

� � ��

1 b1 b

� �D

�……………(1) � D2 =

2b 11 b��

……………(2)

from equation (1) & (2)

� � ��

22 1 bb 11 b 1 b

§ ·� �� ¨ ¸¨ ¸� �© ¹

� � �

� ��

22

2

1 bb 11 b 1 b

��

� �

� b2 + 2b + 1 = b2 + 1 – b3 – b � b3 + 3b = 0 � b(b2 + 3) = 0 � b = 0, b 3i r all are accepted

9. Dx2 – x + D = 0

� |x1 – x2| < 1 � (x1 + x2)2 – 4 x1x2 < 1

� 21 4 1§ · � �¨ ¸D© ¹

� 21 5 0� �D

x1

x2

QUADRATIC EQUATION

242

242


242

� 1–5D2 < 0 � D2 > 15

� 15

§ ·D �¨ ¸© ¹

15

§ ·D �¨ ¸© ¹

> 0 � a � 1 1, ,5 5�§ · § ·

�f � f¨ ¸ ¨ ¸© ¹ © ¹

10. 6 12�S �S

� T � , x2 – 2x secT + 1 = 0

x2 + 2x tanT – 1 = 0

� Sum of roots.� � D1 + E1 = 2secT & D2 + E2 = –2tanT Product of roots � D1E1 = 1 & D2 E2 = –1

D1 , E1 =

22sec 4sec 42

Tr T� 6 12

�S �Sd T d

= 2sec 2 tan

2� Tr T

D1 > E1

D1 , E1 = secT r tanT D1 = secT – tanT

E1 = secT + tanT

� D2 , E2 = 22 tan 4 tan 4

2� Tr T�

= 2 tan 2sec

2� Tr T

= –tanT r secT

6 12�S �S

d T d D2 = – tanT + secT, E2 = – tanT – secT

' D2 > E2 � D2 + E2 = secT – tanT – tanT – secT = – 2tanT

Paragraph for Q.No. 11 to 12

x2 – x + 1, D z E For n = 0, 1, 2.. � D2 = D + 1, E2 =E + 1, an = pDn + qEn a11 + a10 = pD11 + qE11 + pD10 + qE10

= pD10 (D +1) + qE10 (E + 1) = pD10 D2 + qE10 .E2 = pD12 + qE12 = q12

q12 = q11 + q10 an+2 = an +1 + an

D2

E2

D1

E1

QUADRATIC EQUATION

243

243


243

� 1 52�

D , 1 5

2�

E

a4 = a3 + a2 = a2 + a1 + a1 + a0 = a1 + a0 + 2a1 + a0 = 3a1 + 2a0 = 3(pD + qE) + 2(p + q)

� �41 5 1 5a 3 p q 2 p q

2 2§ ·§ · § ·� �

� � �¨ ¸¨ ¸ ¨ ¸¨ ¸ ¨ ¸¨ ¸© ¹ © ¹© ¹

28 = 3p 3 5 3q 3 5p q 2p 2q2 2 2 2� � � � �

because p & q are integers (rational) so not equal to (irrational) only possibility p = q

28 = 7p 7q2 2�

p + q = 8 p = q = 4 p + 2q = 4 + 8 = 12

EXERCISE # 5

1. x2 – 3x + y + 2 = 0 � y = –x2 + 3x – 2

max

Dy k4a�

�

� � �� 9 4 1 2

k4

� � � ��

�

� 1 k4�

2. D < 0

� 4b2 + 20ac < 0 � b2 + 5ac < 0 for equation to be true, a & c have opposite sign ' 4a + 4b + 4c > 9c � 4a + 4b – 5c > 0 � for make it definitely positive. a > 0, b > 0, c < 0

3. Ramesh reads sum of roots correctly = 10 Mahesh reaeds product of roots correctly = –11 Hence required equation is x2 – 10x + 11 = 0 � (x – 11)(x + 1) = 0 � x = 11, –1

y = k

o

QUADRATIC EQUATION

244

244


244

4. 4x2 – 20kx + ( 25k2 + 15k – 66) = 0 condition (1), D t�� 400k2 – 4 × 4 (25k2 + 15k – 66) t 0 � 400k2 – 400k2 – 16 (15k – 66) t 0 � 15k – 66 ≤ 0

� k ≤ 225

………….(1)

condition (2), f (2) > O

16 – 40k + 25k2 + 15k – 66 > 0

� 25k2 – 25k – 50 > 0 � k2 – k – 2 > 0 � (k – 2) (k + 1) > 0 ………….(2)

condition (3), B

2A�

< 2 � 20k

8 < 2 � k <

45

…………(3)

� � x � (–f, –1)

5. Let p(x) = kx3 + 2k2x2 + k3

, (x–2) is a factor � p(2) = 0 � 8k + 8k2 + k3 = 0 � k (8 + 8k + k2) = 0 � k = 0, k2 + 8k + 8 = 0

� k = 8 32 4 82

� r � r

� k1 = 0, k2 4 8 � � , k3 4 8 � �

� k1 + k2 + k3 = – 8 More than on correct 6. f(x) = x2 + bx +c and f (2+ t) = f(2 – t) � real number t.

if t = 1 if t = 2 � f (3) = f(1) � f(4) = f(0) � 9 + 3b + c = 1 + b + c � 16 + 4b + c = c � 8 = –2b � b = –4 � b = –4

f(x) = x2 – 4x + c upward parabola vertex b2a

§ ·�¨ ¸© ¹

= 42

= 2

2 –1 + – +

2

–1

from equation (1), (2) & (3)

2

(1) (3)

(2) 45

225

(2,0)

At same distance from here. (we attain same value)

symmetric

QUADRATIC EQUATION

245

245


245

from fig. we can say that (B) & (D) are correct.

7. x � [1,5] , y = x2 – 5x + 3 f(1) = 1– 5 + 3 = –1 f(5) = 25 – 25 + 3 = 3

� � �25 12D 13 3.254a 4 4

� ��

� b 5 2.5

2a 2�

x � [1, 5] from fig. we can say that (B) & (C) are correct.

8. y = ax2 + bx + c

a > 0

� b 0

2a�

! ��b < 0

D > 0 distinct & real roots � f(0) < 0 � C < 0

9. f(x) = ax2 + bx + c

� a < 0 f(0) > 0 � c > 0

from fig. we can say (A), (B), (C) & (D) all are correct

10. 3 22x 1 0

2x 3x x�

!� �

� � �2

2x 1 0x 2x 3x 1

�!

� � �

� �� 2x 1 0

x x 1 2x 1�

!� �

Option (A) & (D) are correct

11. ax2 + bx +c = 0 (a > 0)

sum of roots sec2T + cosec2T = b

a�

� 1+ tan2T��cot2T�� b

a�

��

0 –1 + – + – +

12� 1

2

2 3 1.5 2.1 2 4 1

2.5 1.5 1

3

–1 –(3.25)

5

D x E

y

–1 o 1

QUADRATIC EQUATION

246

246


246

Product of roots sec2T cosec2T = ca

c, 0a! ��c > 0

�� 1+ tan2T��cot2T�� ca

� 1+ tan2T��cot2T�� ca

� b

a�

= ca

� c + b = 0 ………… (A) is correct ' roots are real so, b2 – 4ac t 0 ……………(B) is correct

� D 0

4a�

d

� � �2b 4ac

04a

� �d from (A) b = – c

� � �2c 4ac

04a

� �d ��

� �c 4a c0

4a�

d � 4a – c ≤ 0 � c t 4a (C) is correct

� 4a – c ≤ 0 and c = –b �� 4a + b ≤ 0 (D) is incorrect

12. y = ax2 + bx + c (a, b, c � R) (A) Product of roots of the corresponding quadratic equation is positive. (a +ib) (a – ib) = a2 + b2 (B) D < 0 no real root. (C) Nothing definite can be said about the sum of the roots, whether positive, negative or �zero.

b 02a�§ · ¨ ¸

© ¹ according to graph

b 0a sum of roots = 0

� (D) D ,E = 2b b 4ac

2a� r �

� D + E = – b 0a ' vertex at y axis b 0

2a�§ · ¨ ¸

© ¹

' a > 0 ? b = 0 D < 0 b2 – 4ac < 0 ��ac > 0 � c > 0 Both roots of the quadratic equation � b = 0 � ax2 + c = 0

y

o x b D,

2a 4a� �§ ·

¨ ¸© ¹

QUADRATIC EQUATION

247

247


247

�� 2 cxa�

�� cx ia

r Purely imaginary

Comprehension Type

13. f(x) = ax2 + bx + c ' A (AB) = 2 , A (AC) = 3 � b2 – 4ac = 4

� b 3

2a�

� & D 2

4a�

��

1a 2

� b = 6a � b = 3 � b2 – 4ac = –4

� 9 – 4 × 12

× c = – 4

� c = 132

� a + b + c = 1 1332 2� � = 10

14. b a cD � �

= 3 + 1 132 2�

= 3 + 7 �� E = 3 – 7 (x – D) (x – E) = 0 x2 – ( D + E) x + DE = 0 x2 – 6x + 2 = 0

15. Range of g(x) = � �21 1a x b 2 x c2 2

§ · § ·� � � � �¨ ¸ ¨ ¸© ¹ © ¹

for x � [–4, 0]

g(x) = x2 + 5x – 6 upward parabola.

� vertexb 5x

2a 2� �

� �

min25 24D 49y

4a 4 4� ��

�

� > @vertexx 4,0� � g (0) = – 6 g ( –4) = –10

g(x) � 49 , 64�ª º�« »¬ ¼

C A

B

3 2

–4�

52

��

0�

(0,–6)� (–4,–10)�

� �5 49,2 4� �

�

QUADRATIC EQUATION

248

248


248

EXERCISE # 6

1. smallest natural number ‘b’

� x2 + 2(a + b)x + (a – b + 8) = 0 has unequal real roots � a � R D > 0 � 4 (a + b)2 – 4(a – b + 8) > 0 � (a +b)2 – (a – b + 8) > 0 � a2 + b2 + 2ab – a + b – 8 > 0 � a2 +(2b– 1)a + (b2 + b – 8) > 0 quadratic in a � (2b–1)2 – 4(b2 + b – 8) < 0 � 4b2 – 4b + 1 – 4b2 – 4b + 32 < 0 � 8b > 33

� b >338

= 4.12 � smallest natural no. b = 5

2. x2 + bx + c = 0 bx2 + cx + 1 = 0 have a common root prove that either b + c + 1 = 0 � � b2 + c2 + 1 = bc + b + c

� (b + c + 1) (b2 +c2 +1 – bc –b – c) = 0 � b3 + bc2 + b – b2c – b2 – bc + b2c + c3 + c – bc2 – bc – c2 + b2 + c2 + 1 – bc –b –c = 0 � b3 – 3bc + c3 + 1 = 0 ……………….(1) � D2 + bD + c = 0 � bD2 + cD + 1 = 0 , if one root is common.

� 2

b cc 1

D = 1 cb 1

�D =

11 bb c

� 2

2 21

bc 1b c c bD D

��

� 2

22

b cc b

§ ·�D ¨ ¸�© ¹

2bc 1c b

�§ ·D ¨ ¸�© ¹

� � ��

2

22

bc 1

c b

�

� =

� ��

2

2

b c

c b

�

�

� b2c2 + 1 – 2bc = (b – c2) (c – b2) � b2c2 + 1 – 2bc = bc – c3 – b3 + b2c2 � b3 + c3 – 3bc + 1 = 0……………….(2) � from equation (1) & (2) Hence proved that one common root only if b + c + 1 = 0 or b2 + c2 + 1 = bc + b + c

QUADRATIC EQUATION

249

249


249

3. P(x) = 4x2 + 6x + 4

= 23 92x 4

2 4

ª º§ ·� � �« »¨ ¸© ¹« »¬ ¼

= 23 72x

2 4

ª º§ ·� �« »¨ ¸© ¹« »¬ ¼

….(1)

Q(x) = 4y2 – 12y + 25

= � �22y 3 25 9ª º� � �¬ ¼

= � �22y 3 16ª º� �¬ ¼ ….(2)

P(x). Q(y) = 28

� 23 72x

2 4

ª º§ ·� �« »¨ ¸© ¹« »¬ ¼

� �22y 3 16ª º� �¬ ¼ = � �2

232x 2y 32

§ ·� �¨ ¸© ¹

+ � �27 2y 34

� + 2316 2x 28

2§ ·� �¨ ¸© ¹

for making product equal to 28.

� 2x + 32

= 0 � x = 3

4�

� 2y – 3 = 0 � y = 32

� unique pair 3 3,4 2�§ ·

¨ ¸© ¹

4. � x2 + 18x + 30 = 22 x 18x 45� � � y + 30 = 2 y 45� � y2 + 900 + 60y = 4(y + 45) � y2 + 56y + 720= 0 � (y + 36) (y + 20) = 0 � y = –36 or y = –20 � x2 + 18x = – 20 x2 +18x = –36 � x2 + 18x + 20 = 0 x2 +18x + 36 = 0 (accepted) (rejected) so only two roots are possible L.H.S. (x2 + 18x + 36)– 6 = – 6

22 x 18x 45� � z negative then product of roots x1x2 = 20

5. a = ? 2

2x ax 23 2x x 1

p

§ ·� �� ¨ ¸� �© ¹

� x � R

� D < 0 means. Positive

QUADRATIC EQUATION

250

250


250

�� 2 23 x x 1 x ax 2� � � � � � � �22x 2x 2� � � � 4x2 + (3 + a) x +1 > 0 x2 + (2 – a)x + 4 > 0 � D < 0 D < 0 � (3+ a)2 – 4 × 4 < 0 (2 – a)2 –16 < 0 � a2 + 6a – 7 < 0 (a – 2)2 – 42 < 0 � (a + 7) (a – 1) < 0……….(1) (a + 2) (a – 6) < 0 ………(2) from equation (1) & (2)

� –2 < a < 1 � a � (–2, 1)

6. a & b are positive

prove that 1 1 1 0x x a x b� �

� � has real roots one between a

3 & 2a3

and the other between 2b3

� & b3

�

�� (x – a) (x + b) + x(x + b) + x(x – a) = 0 � x2 – ax + bx – ab + x2 + bx + x2 – ax = 0 � 3x2 + 2(b – a)x – ab = 0 two real roots D = � �2

Positive

4 b a�� + NPositive12ab > 0

two real roots

� � �af 3 . � �2af 3 = � �2a a3 2 b a ab

9 3ª º§ ·

� � �« »¨ ¸© ¹¬ ¼

� �24a 2a3 2 b a ab

9 3ª º§ ·u � � �« »¨ ¸

© ¹¬ ¼

= 2 2a 2ab 2a ab

3 3 3ª º

� � �« »¬ ¼

2 24a 4ab 4a ab

3 3 3ª º

� � �« »¬ ¼

= 2a ab ab

3 3 3§ ·§ ·� �¨ ¸¨ ¸

© ¹© ¹

= N

2

PositivePositive

a ab ab 03 3 3

§ · § ·� �¨ ¸ ¨ ¸© ¹© ¹��

ensures that one root between a3 & 2a

3

Similarly

2b3

� � �

–7 6 –2 1

QUADRATIC EQUATION

251

251


251

� 2b bf f3 3� �§ · § ·

¨ ¸ ¨ ¸© ¹ © ¹

= � �24b 2b3 2 b a ab

9 3§ ·�§ ·u � � �¨ ¸¨ ¸

© ¹© ¹� �

2b b3 2 b a ab9 3

§ ·�§ ·u � � �¨ ¸¨ ¸© ¹© ¹

= 2 24b 4b 4ab ab

3 3 3§ ·

� � �¨ ¸© ¹

2 2b 2b 2ab ab

3 3 3§ ·

� � �¨ ¸© ¹

= 2ab b ab

3 3 3§ ·�§ · �¨ ¸¨ ¸

© ¹© ¹

= � �� 2b ab ab

9

� �< 0 Hence proved

7. f(y) = y2 + my + 2 divided by y – 1, remainder R1 � f(1) = 1 + m + 2 = m + 3 = R1 ………………..(1) f(y) = y2 + my + 2 divided by y +1, remainder R2 � f(–1) = 1 – m + 2 = 3 – m = R2 ………………..(2) equation (1) – equation (2) � 2m = R1 – R2 = 0 ' R1 = R2 given. � m = 0

8. x2 – ax + b = 0 are real & differ by a quantity which is less than c(c > 0) prove that b

lies between 14 (a2 – c2) & 21 a

4§ ·¨ ¸© ¹

.

� x2 – ax + b = 0

� 2a a 4b

2r �

�

� 2a a 4b

2� � –

2a a 4b2

� � < c

� 2a 4b c� � a2 – 4b t 0 � a2 – 4b < c2 a2 t 4b � a2 – c2 < 4b

� � �2 21 a c b4

� � ……..(1) 2a b

4§ ·

t¨ ¸© ¹

………….(2)

from equation (1) & (2), we can say that � �2

2 21 aa c b4 4

� � d

check for boundary

� 2ab

4 then, x2 – ax +

2a 04

� 4x2 – 4ax + a2 = 0

� so 2ab

4 (not acceptable) ' (2x – a)2 = 0 No differ roots

2a a 4b2

� �

2a a 4b2

� �

QUADRATIC EQUATION

252

252


252

� so � �2

2 21 aa c b4 4

§ ·� � � ¨ ¸

© ¹ Hence proved

9. Let x2 +y2 + xy + 1 t a (x + y) � x, y � R

� x2 + x(y –a) + y2 –ay + 1 t 0 D d 0 � (y – a)2 – 4(y2 – ay + 1) d 0 � y2 – 2ay + a2 – 4y2 + 4ay – 4 d�� –3y2 + 2ay + a2 – 4 d 0 � 3y2 – 2ay + 4 – a2 t 0 D d�� 4a2 – 4 × 3 × (4 –a2) d 0 � a2 – 12 + 3a2 �d�0 � 4a2 – 12 d�� a2 – 3 d 0

� � �� a 3 a 3 0� � d

�

So possible integral solutions. = {–1, 0, 1}

10. (x – D) (x – 4 + E) + (x – 2 + D) (x + 2 – E) = 0

� (x – D) (x – 4 +E) + (x – 2 + D) (x + 2 –E) = 2(x – p) (x – q)………..(1) leading coefficient using comparison = 2 � 2(x –p) (x – q) – (x – D) (x – 4 + E) = 0 from equation (1) � 2(x –p) (x – q) – (x – D) (x – 4 + E) = (x – 2 + D) (x + 2– E) ………… (2)

from equation (1) � 2(x –p) (x – q) – (x – 2 + D) (x + 2 – E ) = (x – D) (x – 4 + E) ………… (3)

absolute value of sum of the roots of the equation (2) & (3) = |2 – D + E – 2 + D + 4 –E| = 4

11. f(x) = x3 + px2 + qx + 72 x2 + ax + b = 0

1

D = b

D 4–E

(2– D) (E– 2)

p

q

3� –1.71

+ – + 3

+1.17

QUADRATIC EQUATION

253

253


253

x2 + bx + a = 0

�� x2 + ax + b = 0 & x2 + bx + a = 0 contains one common root. � D2 + aD + b = 0 � D2 + bD + a = 0

� 2

2 21

a b b aD �D

� �D �E

� D = 1

�� & D . E. 1 = –72

��ab = 72

�� p (x) = x2 + ax + b

� put x = 1 �� p(1) = 0 � 1 + a + b = 0 � 1 + D + E = 1 + b + a = 0

� 1 + b + a = 0

� 1 + b 72b

� = 0

� b + b2 – 72 = 0 � b2 + b – 72 = 0 � (b + 9) (b – 8) = 0 � b = –9, a = 8 or b = 8, a = 9 12 + D2 + E2 = 1 + 81 + 64 = 146

12. f(x) = (a –2)x2 + 2ax + a + 3 lie on the interval (–2, 1) ? � D t 0

� 4a2 – 4(a – 2) (a + 3) t 0 � a2 – (a – 2) (a + 3) t 0 � a2 – (a2 + a – 6) t 0 � –a + 6 t 0 � a d 6 ……..(1) � �case – (1) a > 2 � f(–2) > 0 f(1) > 0 � 4(a –2) –4a + a + 3 > 0 (a –2) + 2a + a + 3 > 0 � a – 5 > 0 ……….(2) 4a + 1 > 0 ………….(3) from equation (1), (2) & (3) � a � (5, 6] case – (2) a < 2, f (–2) < 0, f(1) < 0 we get, a � (–f, 1

4� )

case – (3) a = 2

1

D

1

E = a

–2� 1�

–2� 1�

QUADRATIC EQUATION

254

254


254

� 0.x2 + 4x + 5 = 0

� 5x

4�

, a = 2 (accepted)

From case (1), (2) & (3) we get, ^ ` � @1a , 2 5,64�§ ·� �f � �¨ ¸

© ¹

13. Let D, E¸ J be distinct real numbers such that aD2 + bD + c = (sinT)D2 + (cosT)D aE2 + bE + c = (sinT)E2 + (cosT)E aJ2 + bJ + c = (sinT)J2 + (cosT)J where (a, b, c � R)

� � � ��

2

2

2

a sin b cos c 0

a sin b cos c 0

a sin b cos c 0

½� T D � � T D � °°� T E � � T E� ¾°

� T J � � T J � °¿

these are identities so satisfied for all the values.

� a – sinT = 0 ��a = sinT � b – cosT = 0 ��b = cosT � c = 0

(a) Maximum value of expression = 2 2

2 2a b

a 3ab 5b�

� �

y = 2 2

2 2sin cos

sin 3sin cos 5cosT� T

T� T T� T

= 21

1 3sin cos 4cos� T T� T

= � �

131 sin 2 2 cos 2 12

� T� T�

= 1

33 sin 2 2cos 22

§ ·� T� T¨ ¸© ¹

for maximum, 3 sin 2 2cos 22

§ ·T� T¨ ¸© ¹

must be minimum

� 9 94, 44 4

ª º� � �« »¬ ¼

� 5 5,2 2�ª º« »¬ ¼

� max1y 253

2

�

QUADRATIC EQUATION

255

255


255

(b) �1V ai bj ck � �JJG � � makes an angle

3S

with

� �2V i j 2k � �JJG � � T = ?

� o1 2 1 2V V V V cos60 JJGJJG JJG JJG

� 2 2 2 oa b 2c a b c . 1 1 2 cos60� � � � � �

� sinT + cosT = 2 2 osin cos . 4 cos60T� T � sinT + cosT = o4 cos60

� sinT + cosT = 142

u

� sinT + cosT = 1 � �θ = 2nπ or 4n 12S§ ·�¨ ¸

© ¹

� between T � [0, 2S] for three difference ^ `0, ,22

ST� S

14. 2

2x ax byx 2x 3

� �

� � is [–5, 4] a, b � N (a2 + b2) = ?

� x2(y – 1) + x(2y –a) + 3y – b = 0 ……………….(1) difference with respected to x.

� 2x( y– 1) + x2 dydx

§ ·¨ ¸© ¹

+ (2y –a) + x 2dydx

§ ·¨ ¸© ¹

+ 3dydx

= 0

� (x2 + 2x + 3) dydx

+ 2x(y – 1) – a + 2y = 0

for maxima or minima, dy 0dx

� 2x( y – 1) + 2y – a = 0

� � �

a 2yx2 y 1�

�

………..(2)

Put (2) in (1)

� � ��

� �2

2

a 2yy 1

4 y 1

��

� + � �

� � � �a 2y2y a

2 y 1�

��

+ 3y – b = 0

� (a –2y)2 + 2(a – 2y) (2y – a) + 4(y –1) (3y – b) = 0 � (a – 2y)2 – 2(a – 2y)2 + 4(y –1) (3y – b) = 0 (a – 2y)2 = 4(y –1) (3y – b) ……………………..(3) this has –5 & 4 as roots Put y = –5 � (a + 10)2 = 4(–6) (–15–b) � (a + 10)2 = 360 + 24b ……………(4) Put y = 4 � (a – 8)2 = 144 – 12b ……………(5) equation (4) + equation (5) × 2

④

✓

solutions quadratic equation exercise # 1

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