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qwertyuioop · Jun 14, 2012, 05:50 AM

solve x^2-10x+7=0
what I have so far is this..
x^2-10x____=-7
I don't know what 2 do next!

ebaines · Jun 14, 2012, 06:14 AM

Are you familiar with the quadratic equation? If $ax^2 +bx+c=0$ , then

$<br /> x = \frac {-b \pm \sqrt{b^2-4ac}}{2a}<br />$

qwertyuioop · Jun 14, 2012, 06:18 AM

yes
so then I plug it in and..
x= -10x -+ squre root of 10x^2-4ax^2+7
over 2(-10)

ebaines · Jun 14, 2012, 07:33 AM

Not quite. Given x^2-10x + 7=0, the coefficient of the x^2 term is 'a,' so a=1. The coefficient of the x term is 'b,' so b=-10. And the constant at the end is c, so c=7. Plug these values into the quadratic equation to find values for x.

qwertyuioop · Jun 14, 2012, 07:36 AM

I don't get that!

ebaines · Jun 14, 2012, 07:54 AM

I'll do a different equation for you as an example. Suppose you want to find values of x that satisfy 2x^2-4x-6=0. In this equation a=2, b=-4, and c=-6. So the quadratic equation becomes:

$<br /> x = \frac {-b \pm \sqrt{b^2-4ac}}{2a} = \frac {-(-4) \pm \sqrt{(-4)^2-4(2)(-6)}}{2(2)} = \frac {4 \pm \sqrt{16+48}}{4} = \frac {4 \pm 8} 4 = 1 \pm 2.<br />$

So the answer is x equals either 3 or -1. To check that this is correct it's a good idea to plug these two values back into the original equation and see if it works:

For x=3:
2(3)^2 -4(3)-6 = 18-12-6 = 0

For x=-1:
2(-1)^2 -4(-1) -6 = 2+4-6 = 0

So these values of x do indeed satisfy the original equation.

You can apply this same process to your problem. Hope this helps.

qwertyuioop · Jun 14, 2012, 07:57 AM

Yes it does thanks! <3