Question

Solve x2 – 6x = 27 by completing the square

Answer 1

x^2-6x-27=0
(x-9)(x+3)=0
x=9;x=-3

I guess

Answer 2

@mae23 that's not completing the square...correct answers, but wrong method.

\[x^2-6x=27\]

Our goal is to turn the left hand side into something that looks like \((x+a)^2\) for some suitable value of \(a\). It turns out that the suitable value of \(a\) will be 1/2 of the coefficient of \(x\).

\[(x-\frac{1}{2}*6)^2 = (x-3)^2 = x^2-3x-3x+9=x^2-6x+9\]

We can't just write that on the left side, however, because there's no 9 there. What we can do is add 9 to both sides:
\[x^2-6x+9 = 27+9\]Now we can rewrite the left hand side as \[(x+3)^2 = 27+9\]\[(x+3)^2 = 36\]Take the square root of each side:
\[\sqrt{(x+3)^2} = \sqrt{36}\]\[(x+3) = \pm 6\]Now if you solve that very simple equation for the two values of \(x\) that make it true, you've got your answer.

Answer 3

Now, a bit of explanation as to why this works:

\[(x+a)^2 = (x+a)(x+a) = x(x+a) + a(x+a) = x*x + x*a + a*x + a*a\]\[\qquad=x^2+2ax + a^2\]

That means if we have \[x^2 + 2ax\](and with the right choice of \(a\), that will cover any case we need), we can make it a perfect square (aka "complete the square") by adding \(a^2\) and rewriting as \((x+a)^2\).

However, if we don't have some constant floating around on that side of the equation from which we can subtract the value \(a^2\), we'll have to add \(a^2\) to both sides of the equation to preserve the equality.

Answer 4

factoring is easier than completing the square. sorry about that

Answer 5

Sometimes you'll need to do this technique multiple times on the same equation. For example, in expanded form, the equation of a circle might look like this:

\[1-2x+x^2-4y+y^2=0\]Let's regroup that a bit:\[(x^2-2x) + (y^2-4y) + 1 = 0\]Let's complete the square on \(x\):
\[(x^2-2x+(-2/1)^2 ) + (y^2-4y) + 1 = 0 + (-2/1)^2\]\[(x^2-2x+1) + (y^2-4y) + 1 = 0+1\]Now we can rewrite the leftmost grouping as a perfect square:\[(x-1)^2 + (y^2-4y)+1=1\]Now do the same for \(y\):\[(x-1)^2 + (y^2-4y+(-4/2)^2) + 1 = 1 + (-4/2)^2\]\[(x-1)^2 + (y^2-4y+4) + 1 = 1 + 4\]\[(x-1)^2 + (y-2)^2 + 1 = 5\]\[(x-1)^2 + (y-2)^2 = 4\]So that turned out to be the equation of a circle of radius \(r = \sqrt{4} = 2\) with center \((1,2)\)

Answer 6

@mae23 solve \[y = ax^2 + bx + c\] by factoring :-)

Answer 7

@whpalmer4 cool your so smart. u should help me too.lol