Equations in Quadratic Form

The "u" Substitution Method

045 24 xxBefore we solve the above equation, let's solve a quadratic equation that we know how to solve.

0452 uu Factor

014 uu Set each factor = 0 and solve

1,4 uuLet's use this to solve the original equation by letting u = x2.

045 24 xx

0452 uu Factor

014 uu Set each factor = 0 and solve

1,4 uuNow that we've solved for u we have to re-substitute to get x back. Remember u = x2 so let's substitute.

If u = x2 then square both sides and get u2 = x4. Substitute u and u2 for x2 and x4.

1,4 22 xx

Solve for x by square-rooting both sides and don't forget the 1,2 xx

044 4

1

2

1

zz

0442 uu

022 uu Factor & set each factor = 0 and solve2u

You can determine if an equation is of quadratic form where you can use the "u" substitution method if you call the middle variable and power u and then square it and get the first term's variable and power.

So let u = z1/4 and get u2 = z1/2. Substitute u and u2 for z1/4 and z1/2.

24

1

zuSolve for z by raising both sides to the 4th power

444

1

)2()( z

2

12)(

4

1

zz

16z

087 36 xx

0872 uu

018 uu Factor & set each factor = 0 and solve

1,8 uu

Let's try one more. Call the middle variable u and then square it to see if you get the first term's variable.

So let u = x3 and get u2 = x6. Substitute u and u2 for x3 and x6.

Solve for x by taking the cube root of both sides

623)( xx

1,2 xx

1,8 33 xx

Acknowledgement

I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this PowerPoint.

www.slcc.edu

Shawna has kindly given permission for this resource to be downloaded from www.mathxtc.com and for it to be modified to suit the Western Australian Mathematics Curriculum.

Stephen CorcoranHead of MathematicsSt Stephen’s School – Carramarwww.ststephens.wa.edu.au