Question

solve by completing the square: x^2+3x=88

Answer 1

Do you understand the idea behind completing the square, or do you want the whole explanation?

Answer 2

I understand completing the square i just don't get how it works with this equation.

Answer 3

Hm, subtract both sides 88, what do you get?

Answer 4

x^2+3x-88=0

Answer 5

\[x^2+3x =88\]Take \(2a=3\) (where \(3\) is the coefficient of the \(x\) term. Then \(a = 3/2\), so that your completed square will be \((x+a)^2 = (x+3/2)^2\)

To complete the square, you need to square \(a\) and add it to both sides.
\[x^2+3x=88\]\[(x+3/2)^2 = 88+(\frac{3}{2})^2\]
Note that we already took the piece we added on the left side and incorporated it into \((x+3/2)^2\).

\[(x+3/2)^2 = \frac{361}{4}\]\[(x+3/2) = \pm\sqrt{\frac{361}{4}}\]
You can do the rest, I'm sure...

Answer 6

Yes, take the "x" term and divide it by 2 and square it. Then put it back in the parenthesis.

\(\ \sf ax^2 + \color{blue}{b}x - c = 0 \)

Here the "x" value is 3, so \(\ \dfrac{3}{2} \) by the way, never simplify a fraction because it's really hard to factor a decimal, if it simplifies into a whole number then yes, do it. \(\ \sf \dfrac{3}{2} = (\dfrac{3}{2})^2 = \dfrac{3^2}{2^2} = \dfrac{9}{4} \)

Now plug that 9/4 back into (x^2 + 3x + ___) - 88 = 0 ... (x^2 + 3x + 9/4) - 88 = 0

Because you added a 9/4 you have to subtract it from -88 to keep the equation balanced.

\(\ \sf \Large (x^2 + 3x + \dfrac{9}{4}) - 88 -\dfrac{9}{4} = 0 \)

\(\ \sf \Large (x^2 + 3x + \dfrac{9}{4}) -\dfrac{361}{4} = 0 \)

The only thing that you need to now is factor (x^2 + 3x + 9/4), a quick way to do it is by simple factoring the second term, it will \(\ \sf always\) be the second term divided by 2. So you would get \(\ \sf (x + \dfrac{b}{2})^2 - 361/4 = 0\), recall that our b/2 here is 3/2. So you would get (x+3/2)^2 - 361/4 = 0.

Now just add \(\ \sf -\dfrac{\color{blue}{361}}{\color{blue}{4}}\) when you add -361/4 to both sides you get

Now the next step is to solve for "x", so just do the inverse of everything! Square root both sides, \(\ \sf \sqrt{\dfrac{361}{4}} = ? \) .. simply square root 361 and square root 4 >.<

Answer 7

You get \(\ \sf (x^2 + 3x + 9/4) = 361/4 \) I don't know why it didn't come out :<

Answer 8

thanks!

Answer 9

Yw, what did you get :O

Answer 10

(x^2 + 3x + 9/4) = 361/4 isn't the full answer, I just put it back because LaTeX didn't show it in my previous post. :P

Answer 11

it comes out to 8,-11

Answer 12

i think hold on let me check again

Answer 13

Yes, \[x = 8, x=-11\]is correct

Answer 14

When you're trying to solve for x and you're going to square root, I forgot to add the \(\ \pm \)

\(\ \sf \Large \sqrt{(x + 3/2)^2} = \pm \sqrt{\dfrac{361}{4}} \)

Yes you are correct, x = {-11,8}! When you plug in 8 or -11 for x you get 88 :>

Answer 15

yeah! thanks so much!

Answer 16

Np ;p