# Math Help - completing square, is this correct??

1. ## completing square, is this correct??

hi, i need to completing square on this:

x^2+2x-5=0

for this would i be correct if i got this:

(x+1)^2 -1 -5= (x+1)^2-6 ????

if this isn't correct where am i going wrong?

thanks!

2. You're on the right track.

so,

$x^2+2x-5=0$

$(x+1)^2 -1 -5= 0$

$(x+1)^2-6 =0$

$(x+1)^2=6$

$x+1=\pm{\sqrt{6}}$

now you can find x

3. thanks...! however in class we have never been told to find "x" in these questions we just leave it as how i left it in post #1, thanks again!

4. if you are to complete the square then that generally implies you are to solve for x. completing the square is a type of factoring. when you factor it is to find solutions. if you need to graph the function then you would not take the square root and keep it in standard form.

standard form for graphing:

$f(x)=(x+1)^2-6$

general form:

$x^2+2x-5=0$

5. Originally Posted by skoker
if you are to complete the square then that generally implies you are to solve for x. completing the square is a type of factoring. when you factor it is to find solutions. if you need to graph the function then you would not take the square root and keep it in standard form.

standard form for graphing:

$f(x)=(x+1)^2+6$
If this was the original problem, it should be $(x+1)^2- 6$

I would NOT say that "if you are to complete the square then that generally implies you are to solve for x." All it implies is that you are to complete the square. I also would not say "when you factor it is to find solutions". There may be many reasons to factor a polynomial.

general form:

$x^2+2x-5=0$

6. There are many reasons you might want to complete the square. Some of the more basic ones are to:

(2) Put a nondegenerate conic section into standard form (ellipse, hyperbola, parabola).
(3) Evaluate an integral.

7. yes i actually thought about completing square for conics after i posted. post too soon I suppose...

8. Originally Posted by harish21
You're on the right track.

so,

$x^2+2x-5=0$

$(x+1)^2 -1 -5= 0$

$(x+1)^2-6 =0$

$(x+1)^2=6$

$x+1=\pm{\sqrt{6}}$

now you can find x
hi again, you find x i know i have to take away 1 from each side to get, x= square root 6 -1

im unsure what to do after this because two numbers times together to make this doesn't work because there isn't a square number to do this.

could you let me know how i could solvew this please?

cheers

9. Originally Posted by andyboy179
hi again, you find x i know i have to take away 1 from each side to get, x= square root 6 -1

im unsure what to do after this because two numbers times together to make this doesn't work because there isn't a square number to do this.

could you let me know how i could solvew this please?

cheers
We have

$x = \pm \sqrt{6} - 1$

When x is squared, there are two solutions. Sometimes they are the same one, sometimes they are complex, but there are two. In this case one x is
$\sqrt{6} - 1$
while the other is
$-\sqrt{6} - 1$