Complete the square in the form $(px+q)^2+r, p > 0$

Question

I'm going over some completing the square questions and I need to express, in the form: $(px+q)^2+r, p > 0$

the quadratic equation is $16x^2-8x+11$

I know how to get it in the form $p(x+q)^2+r$

So can someone show me how to get from $ax^2+bx+c$ to $(px+q)^2+r, p > 0$

Answer 1

To get from $ax^2+bx+c$ to $(px+q)^2+r$, you can do the following:- $$\begin{align}ax^2+bx+c&=a\left(x^2+\frac{b}{a}x+\color{red}{\left(\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2}+\frac{c}{a}\right)\\&=a\left(\left(x+\frac{b}{2a}\right)^2+\frac{c}{a}-\frac{b^2}{4a^2}\right)\\&=a\left(x+\frac{b}{2a}\right)^2+\left(c-\frac{b^2}{4a}\right)\\&=\left(\sqrt{a}x+\frac{b}{2\sqrt{a}}\right)^2+\left(c-\frac{b^2}{4a}\right)\end{align}$$ For the particular polynomial in question, just plug in the values ($a=16$, $b=-8$,$c=11$)

Answer 2

You can either do:

$$(px + q)^2 + r = 16x^2 - 8x + 11 \iff p^2x^2 + 2pqx + q^2 + r = 16x^2 - 8x + 11$$ and find the values of $p,q,r$ by identification, or do:

$$16x^2 - 8x + 11 = (4x)^2 - 2(4x)(1) + 11 = (4x)^2 - 2(4x)(1) + 1^2 + 10 = (4x - 1)^2 + 10$$

Hence:

$$p = 4, \ q = -1,\ r = 10$$

Answer 3

$16x^2-8x+11 = 16[x^2-\frac{1}{2}x]+11 = 16[(x-\frac{1}{4})^2-\frac{1}{16}]+11 \\ = 16(x-\frac{1}{4})^2-1+11 = 16(x-\frac{1}{4})^2+10 = (4x-1)^2+10.$

Answer 4

\begin{align*}16x^2 - 8x + 11 &= 16 \left(x^2 - \frac12\right) +11\\ &= 16\left[\left( x- \frac12\right)^2 -\frac14\right] + 11\\ &= 16\left( x- \frac12\right)^2 - 4 + 11\\ &= \left[\sqrt{16} \left(x-\frac12\right)\right]^2 + 7\\ &= \left(4x - 2\right)^2 + 7 \end{align*}