Question

Completing the Square??
x^2+11x-6=0
x^2+11x=6
Having trouble after that...

Answer 1

okay, you're off to a good start. Now take half of the coefficient of the \(x\) term, square it, and add that amount to both sides of the equation. After you do that, you can replace the left hand side with
\[(x+a)^2\]where \(a = \frac{1}{2}\) the coefficient of the \(x\) term.

Answer 2

Thats where i am stuck at the coefficient of 1 and 11? then square it

Answer 3

Here's why you can do this:

If we square \((x+c)\) we get
\[(x+c)(x+c) = x^2 + cx + cx + c^2 = x^2 + 2cx + c^2\]

That means if we have \(x^2 + bx\) we can rewrite it as \((x+\frac{b}{2})^2\) so long as we add \((\frac{b}{2})^2\) to the other side of the equals sign.

Answer 4

Coefficient of \(x\) term is the coefficient of \(11x\). What is that?

Answer 5

is it 121/4

Answer 6

Well, yes, after you take half of it and square it. But I just wanted the coefficient, which is \(11\). Half of that is \(\frac{11}{2}\). We need to remember that value, because we are going to replace the left hand side with \((x+\frac{11}{2})^2\) after we square that quantity and add it to both sides.

Answer 7

so after we take \(11\), divide it by \(2\), square it and add it to both sides, we get:
\[x^2 + 11x + (\frac{11}{2})^2 = 6 + (\frac{11}{2})^2\]

we can rewrite the left hand side as

\[(x+\frac{11}{2})^2 = 6+(\frac{11}{2})^2\]

Answer 8

Now we need to simplify the right hand side. Can you do that? Multiply out the fraction, and put everything on the right hand side over a common denominator.

Answer 9

yes... let me work the problem real quick

Answer 10

is it -121/4+-\[\frac{ -121 }{4 }\pm \sqrt{\frac{ 145 }{4 ? }}\]

Answer 11

take your time, accuracy is important :-)

sadly, your answer does not appear to be correct.

Answer 12

What do you get for \[6+(\frac{11}{2})^2\]

Answer 13

\(\frac{145}{4}\)?

Answer 14

yes

Answer 15

Good.

\[(x+\frac{11}{2})^2 = \frac{145}{4}\]If we take the square root of both sides, we get
\[x+\frac{11}{2} = \pm\sqrt{\frac{145}{4}}\]Right?

Answer 16

yes i see where i messed up at

Answer 17

\[\frac{ -11 }{ 2? }\pm \frac{ \sqrt{145} }{2 ? }\]

Answer 18

\[x = \frac{1}{2}(-11\pm\sqrt{145})\]

Answer 19

or \(x \approx -11.521\) and \(x \approx 0.521\)

Answer 20

thanks for your help

Answer 21

i got it correct

Answer 22

Great!

does the procedure make sense to you now, or would you like to do a couple of practice problems with me?

Answer 23

sorry just saw your message i would like to practice more problems like this when you see this.