How do I complete the square of $y= -4x^2-2x-4$?

Question

$y = -4x^2 - 2x - 4$

I just can't figure this out, do I divide the second number and the third number by $4$ then by $2$ and then add the product to the second one and subtract it from the third one?

Answer 1

First, you need to isolate x^2 from the equation by factor the -4 out, which leave the equation like this:

$$y= -4(x^2+ \frac{1}{2} x ) -4 $$

We know that the equation of square binomials is:

$(x+b)^2= x^2+2xb+b^2$

To write the binomial $x^2 + \frac{1}{2}x$ to a square binomials, we need to find b^2

We got : $2xb = \frac{1}{2} x $
=> $2b= \frac{1}{2}$
=> $b= \frac{1}{4}$
=> $b^2= \frac{1}{16}$

To balance the equation we need to add and subtract b^2 in the parentheses

$$y= -4(x^2+ \frac{1}{2} x +\frac{1}{16} -\frac{1}{16}) -4$$

Distribute the $-\frac{1}{16}$ out and combine like term

$$y= -4(x^2+ \frac{1}{2} x +\frac{1}{16}) +\frac{1}{4} -4$$

$$y= -4(x^2+ \frac{1}{2} x +\frac{1}{16}) -\frac{15}{4}$$

Now convert the trinominal inside of the parentheses to square binomial

$$y= -4(x^2+ 2\frac{1}{4} x +(\frac{1}{4})^2) -\frac{15}{4}$$

$$y= -4(x + \frac{1}{4})^2 -\frac{15}{4}$$

Answer 2

$-4x^2 -2x -4 = -(4x^2 + 2x + 4)$

The $4x^2 +2x + 4$ must come from some $(2x+b)^2$, to get the right square, and this has linear term $4b$, which should equal $2x$ so $b=\frac{1}{2}$. Now $(2x + \frac12)^2 = 4x^2 + 2x + \frac14$, so we need an extra $3\frac34$ to get $4$, like we need. So in all

$$-4x^2 -2x -4 = -\left((2x+\frac12)^2 + 3\frac34\right)$$

Answer 3

There is a general formula you can use it. Every quadratic function can be writen as:

$$f(x)= a(x-p)^2+q$$wher $p=-{b\over 2a}$ and $q= {-D\over 4a}$ and $D=b^2-4ac$.

Answer 4

The simplest strategy is to multiply both sides by $4$: $$ 4y=-(\underbrace{16x^2+8x}_{\text{to complete}}+16) $$ Then notice that $(4x+1)^2=16x^2+8x+1$, so we have $$ 4y=-((4x+1)^2+15) $$ and then $$ y=-\frac{1}{4}(4x+1)^2-\frac{15}{4} $$

In general, if you have a polynomial like $ax^2+bx+c$, you can easily complete the square by multiplying by $4a$: $$ \underbrace{4a^2x^2+4abx}_{\text{to complete}}+4ac= 4a^2x^2+4abx+b^2-b^2+4ac=(2ax+b)^2+(4ac-b^2) $$ Then you can divide back by $4a$. In this particular case, it's not necessary to multiply by $16$, as $4$ suffices.

Answer 5

In general, if you want to complete the square on $ax^2+bx+c$, here's the formula and derivation

Suppose that $$ax^2+bx+c=A(x+B)^2+C$$ Then: $$ax^2+bx+c=Ax^2+2ABx+AB^2+C$$ Therefore: $$A=a$$ $$b=2aB$$ $$c=aB^2+C$$ Therefore: $$B=\frac{b}{2a}$$ $$C=-a(\frac{b}{2a})^2+c=-\frac{b^2}{4a}+c$$ Plug it all in: $$ax^2+bx+c=a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}$$ Which is the complete square.

Answer 6

Method 1:

$y = -4x^2 -2x -4$

$\frac {y}{-4} = x^2 + \frac 12x + 1$

$-\frac {y}{4}- 1 = x^2 + 2*\frac 14 x$

$-1-\frac y4 +(\frac 14)^2= x^2 + 2*\frac 14 x+ (\frac 14)^2$

$-1+\frac 1{16}-\frac y4 = (x + \frac 14)^2$

$(x+ \frac 14)^2 = -\frac {15}{16} - \frac y4 = \frac {-15 - 4y}{16}$

$x + \frac 14 = \pm \sqrt {\frac {-15-4y}{16} }= \frac {\sqrt {-15 - 4y}}4$

$x = \frac {\sqrt {-15 - 4y}}4 -\frac 14 = \frac {\sqrt{-15 - 4y} - 1}4$

Method 2:

$y = -4x^2 - 2x -4 \iff y+4 = -4x^2 -2x$

We want $-4x^2 - 2x$ to be the first to summands of $(mx + n)^2 = m^2x + 2mnx + n^2$. But $m^2$ is positive while $-4$ is negative.

So we multiple everything by $-1$.

$y+4 = -4x^2 -2x \iff -y - 4 = 4x^2 + 2x$. Now we want $4 = m^2$ and $2 = 2mn$.

Or in other words $m = \pm \sqrt 4$ and $n = \frac 1m$. Or in other words $m = \pm 2$ and $n = \frac 12$.

$-y -4 = 4x^2 + 2x = m^2x^2 + 2mnx \iff$

$-y-4 + n^2 = -y -4 + (\frac 12)^2= 4x^2 + 2x + (\frac 12)^2 = m^2x^2 + 2mnx +n^2 = (mx + n)^2 = (2x + \frac 12)^2$

$\iff -y -4 +\frac 14 = -y -\frac {15}4 = (2x + \frac 12)^2$

$\iff 2x +\frac 12 =\pm \sqrt {-y -\frac {15}{4}}$

$\iff 2x =-\frac 12 \pm \sqrt {-y-\frac {15}4}=-\frac 12 \pm \frac {\sqrt{-4y-15}}2 = \frac {-1 \pm\sqrt {-4y - 15}}2$

$\iff x = \frac {-1 \pm \sqrt{-4y -15}}4$