Completing the Square: A Different Approach for Solving Quadratic Equations?

[Mo]

I have only just recently started this topic ... (3 topic in AS maths for me)

I have 1 question with two parts.I just can't seem to get the answer!


"Express, in the form (px+q)^2 + r whereby p > 0"

a) 16x^2 -8x +11
b) 9x^2 +3x +1

I don't seem to find a problem doing any of these when the coefficient of X squared is 1, but when it is bigger than 1, it causes me problems!

I would be gratefull if some once could at least help me through the first 1, so i can understand the method.Thanks.

Regards
Mo

 

[dextercioby]

I have only just recently started this topic ... (3 topic in AS maths for me)

I have 1 question with two parts.I just can't seem to get the answer!


"Express, in the form (px+q)^2 + r whereby p > 0"

a) 16x^2 -8x +11
b) 9x^2 +3x +1

I don't seem to find a problem doing any of these when the coefficient of X squared is 1, but when it is bigger than 1, it causes me problems!

I would be gratefull if some once could at least help me through the first 1, so i can understand the method.Thanks.

Regards
Mo

I'll solve a) and let u take b).
a)16x^{2}=(4x)^{2}
16x^{2}-8x=(4x)^{2}-2\cdot 4x
Then:
16x^2 -8x +11=(4x-1)^{2}+10

Daniel.

 

[shmoe]

A naive way of doing this is to write:

16x^2 -8x +11=(px+q)^2 + r

and expand the right hand side (multiple out the squared part and collect powers of x). Now match the coefficients to solve for p, q, and r.

 

[Mo]

Thanks both for your help!

 

[ComputerGeek]

I'll solve a) and let u take b).
a)16x^{2}=(4x)^{2}
16x^{2}-8x=(4x)^{2}-2\cdot 4x
Then:
16x^2 -8x +11=(4x-1)^{2}+10

Daniel.

I have never seen your method before. I cannot say it makes sense to me. care to explain it?

 

[dextercioby]

I have never seen your method before. I cannot say it makes sense to me. care to explain it?

It's not a bigdeal.
Let's pick an arbitrary polynom of degree 2:ax^{2}+bx+c.
U want to put in the form (px+q)^{2}+r.
The direct method is to equal the two expressions and identify the coefficients of the powers of "x".That what Shmoe said.
I found another method which can be thought of being intuitive,and sometimes useful as well.
Take the square:ax^{2}+bx.It can be put like:
(\sqrt{a} x)^{2}+2\sqrt{a}\frac{b}{2\sqrt{a}} x+\frac{b^{2}}{4a}-\frac{b^{2}}{4a} okay??
You restrain the square and add "c" in both sides to get:
ax^{2}+bx+c =(\sqrt{a} x +\frac{b}{2\sqrt{a}})^{2} +(c-\frac{b^{2}}{4a})

And u can easily find (p,q,r).

Daniel.