aim: the discriminant course: adv. alg, & trig. aim: what is the discriminant and how does it...
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Aim: The Discriminant Course: Adv. Alg, & Trig.
Aim: What is the discriminant and how does it help us determine the roots of a parabola?Do Now: Graph
x2 – 2x – 3 = yx2 – 6x + 7 = yx2 – 4x + 4 = yx2 – 4x + 5 = y
Describe the roots for each.
Aim: The Discriminant Course: Adv. Alg, & Trig.
The Graph, the Roots, & the x-axis
y = ax2 + bx + c Equation of parabola
y = 02 real roots
2 real equalroots
NO real roots,complex
Aim: The Discriminant Course: Adv. Alg, & Trig.
Parabolas
x2 – 4x + 5 = y
x2 – 2x – 3 = y x2 – 6x + 7 = y
x2 – 4x + 4 = y
Imaginary roots
{2 i }
2 real rational roots{-1 and 3}
2 real rational rootsthat are equal {2}
2 real irrational roots
{3 2 }
Aim: The Discriminant Course: Adv. Alg, & Trig.
x2 – 2x – 3 = 0
x ( 2) ( 2)2 4(1)( 3)
2(1)
x ( 2) 4 12
2(1)
x ( 2) 16
2(1){-1 and 3}
Quadratic Formula Solutions
x2 – 6x + 7 = 0
x ( 6) ( 6)2 4(1)(7)
2(1)
x ( 6) 36 28
2(1)
x ( 6) 8
2(1)
{3 2 }
x2 – 4x + 4 = 0
x ( 4) ( 4)2 4(1)(4)
2(1)
{2}
x ( 4) 16 16
2(1)
x ( 4) 0
2(1)
x2 – 4x + 5 = 0
x ( 4) ( 4)2 4(1)(5)
2(1)
{2 i }
x ( 4) 16 20
2(1)
x ( 4) 4
2(1)
Aim: The Discriminant Course: Adv. Alg, & Trig.
The Discriminant Knows!
x2 – 4x + 5 = y
x2 – 2x – 3 = y x2 – 6x + 7 = y
x2 – 4x + 4 = y
Imaginary roots
{2 i }
2 real rational roots{-1 and 3}
2 real rational rootsthat are equal {2}
2 real irrational roots
{3 2 }
discri minant : 16 4
discr : 8 irrational
discr : 0 0
discr : 4 imaginary
Aim: The Discriminant Course: Adv. Alg, & Trig.
The Discriminant
x b b2 4ac
2aThe discriminant -
the expression under the radical sign. Itdetermines the nature of the roots of a quadratic equation when a, b, and c are rational numbers.
b2 – 4ac
Quadratic Formula
x b b2 4ac
2a
Aim: The Discriminant Course: Adv. Alg, & Trig.
The Nature of the Roots - Case 1
x2 – 2x – 3 = y
2 real rational roots{-1 and 3}
b2 – 4ac =
If the b2 – 4ac > 0 and b2 – 4ac is a perfectsquare, then the roots of the equation
ax2 +bx + c = 0 are real, rational and unequal.
(-2)2 – 4(1)(-3)
the discriminant
4 + 12 = 16
the discriminantis a perfect square
a = 1, b = -2, c = -3
Aim: The Discriminant Course: Adv. Alg, & Trig.
The Nature of the Roots - Case 2
b2 – 4ac =
If the b2 – 4ac > 0 and b2 – 4ac is not a perfectsquare, then the roots of the equation
ax2 +bx + c = 0 are real, irrational and unequal.
(-6)2 – 4(1)(7)
the discriminant
36 – 28 = 8
the discriminantis a positive number,but not a perfect squ.
a = 1, b = -6, c = 7 x2 – 6x + 7 = y
2 real irrational roots
{3 2 }
Aim: The Discriminant Course: Adv. Alg, & Trig.
The Nature of the Roots - Case 3
b2 – 4ac =
If the b2 – 4ac = 0, then the roots of the equation ax2 +bx + c = 0 are real, rational and equal.
(-4)2 – 4(1)(4)
the discriminant
16 – 16 = 0
the discriminantis zero
a = 1, b = -4, c = 4 x2 – 4x + 4 = y
2 real rational rootsthat are equal {2}
Aim: The Discriminant Course: Adv. Alg, & Trig.
The Nature of the Roots - Case 4
b2 – 4ac =
If the b2 – 4ac < 0, then the roots of the equation ax2 + bx + c = 0 are imaginary.
(-4)2 – 4(1)(5)
the discriminant
16 – 10 = -4
the discriminantis a negative number
a = 1, b = -4, c = 5 x2 – 4x + 5 = y
Imaginary roots
{2 i }
Aim: The Discriminant Course: Adv. Alg, & Trig.
The Discriminant
Value of Discriminant Nature of roots of
ax2 + bx + c = 0
b2 - 4ac > 0 and b2 - 4ac is a perfect square
real, rational, unequal
b2 - 4ac > 0 and b2 - 4ac is not a perfect square
real, irrational, unequal
b2 - 4ac = 0 real, rational, equal
b2 - 4ac < 0 imaginary
Aim: The Discriminant Course: Adv. Alg, & Trig.
Model Problem
The roots of a quadratic equation are real, rational, and equal when the discriminant is
1) -22) 23) 04) 4
The roots of the equation 2x2 – 4 = 4 are1) real and irrational2) real, rational and equal3) real, rational and unequal4) imaginary
Aim: The Discriminant Course: Adv. Alg, & Trig.
Model Problem
Find the largest integral value of k for whichthe roots of the equation 2x2 + 7x + k = 0 are real.
a = 2, b = 7, c = kIf the roots are real, then the discriminantb2 - 4ac > 0.
substitute into b2 - 4ac ≥ 0 (-7)2 - 4(2)(k) ≥ 0
49 - 8(k) ≥ 0
49 ≥ 8k
6 1/8 ≥ k
The largest integer: k = 6 = c
72 - 4•2•7 = 49 – 56 = -7 check: c = 7
72 - 4•2•6 = 49 – 48 = 1 c = 6
imaginary