completing square, is this correct??

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February 18th 2011, 07:14 AM
andyboy179

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!
February 18th 2011, 07:23 AM
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
February 18th 2011, 07:26 AM
andyboy179

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!
February 19th 2011, 03:32 AM
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:

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

general form:

$x^2+2x-5=0$
February 19th 2011, 03:55 AM
HallsofIvy

Quote:

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.

Quote:

general form:

$x^2+2x-5=0$
February 19th 2011, 04:05 AM
DrSteve

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

(1) Solve a quadratic equation.
(2) Put a nondegenerate conic section into standard form (ellipse, hyperbola, parabola).
(3) Evaluate an integral.
February 19th 2011, 04:31 AM
skoker

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

(Wait)
March 7th 2011, 06:34 AM
andyboy179

Quote:

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
March 7th 2011, 02:17 PM
getsallad

Quote:

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
$-\sqrt{6} - 1$