Friday

December 19, 2014

December 19, 2014

Posted by **q** on Friday, August 3, 2007 at 5:28pm.

solve for x.

thanks so much.

Please keep in mind that

=)

it was a word problem and i already got it into an equation. i tried solving it but having a hard time on where to start to solve for x. Thanks.

Try expanding the part you have in parentheses (use the FOIL method), then combine like terms and work with what's left.

this was what i got.

-2x^2-40

possible to factor? how? please help.

You can factor a two out. However, I didn't get that set of values when I expanded and combined the terms from the original set up you posted.

Like in case of the other problem, if you plan to solve the quadratic using trial and error (using sum and product of roots), then you could try to find one or both roots by inspection or other manipulations that require less work than expanding out the quadratic and writing it in standard form first.

x^2+(x-2)^2 = (2x-6)^2

You don't have to look at this long to see that x = 2 is a solution. To find the other solution, let's write the equation as:

x^2 - (2x-6)^2 = -(x-2)^2

And use the formula

A^2 - B^2 = (A+B)(A-B)

[x + (2x-6)][x - (2x-6)] = -(x-2)^2 --->

(3x - 6)(x+6) = (x-2)^2

We can write 3x-6 = 3(x-2). If we divide both sides by x - 2 we can find the oher slutiuon:

3x + 18 = x - 2 --->

x = 8

Correction (a few sign errors):

And use the formula

A^2 - B^2 = (A+B)(A-B)

[x + (2x-6)][x - (2x-6)] = -(x-2)^2 --->

(3x - 6)(x-6) = (x-2)^2

We can write 3x-6 = 3(x-2). If we divide both sides by x - 2 we can find the oher slutiuon:

3x - 18 = x - 2 --->

x = 8

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