Question 622630
Hi there,
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Solve the equation by factorization method.
{{{ x^2+(x+2)^2=650}}}
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I see one problem: {{{(x+2)^2<>x^2+4}}}. You need to expand it out like this:
{{{(x+2)^2=(x+2)(x+2)}}}
{{{(x+2)^2=x^2+2x+2x+4}}}
{{{(x+2)^2=x^2+4x+4}}}
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You can probably take it from there, but I'll continue on just in case.
{{{x^2+(x+2)^2=650}}}
{{{x^2+(x^2+4x+4)=650}}}
{{{2x^2+4x-646=0}}}
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I find it much easier to factor when the leading coefficient is 1, so I will divide both sides of the equation by 2. (This step is not necessary, but it won't change the answer either.)
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{{{x^2+2x-323=0}}}
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In order to factor this equation, we want two numbers whose product is -323 and whose sum is 2. It turns out that 17*19=323. I figured this out by checking the prime numbers (2, 3, 5, 7, 11, ...) until I found one that worked. The numbers we want are -17 and 19; their product is -323 and their sum is 2.
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{{{x^2+2x-323=(x-17)(x+19)=0}}}
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Since the product of these factors is 0, either {{{x-17=0}}} or {{{x+19=0}}}. 
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Therefore, x=17 or x=-19.
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It's always a good idea to check the answers in the original problem.
x=17:
{{{ x^2+(x+2)^2=650}}}
{{{ (17)^2+(17+2)^2=650}}}
{{{289+361=650}}}
{{{650=650}}} Check!!!
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x=-19
{{{ x^2+(x+2)^2=650}}}
{{{ (-19)^2+(-19+2)^2=650}}}
{{{361+289=650}}}
{{{650=650}}} Check! Check!
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Hope this helps! Feel free to email if you still have questions about this.
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Ms.Figgy
math.in.the.vortex@gmail.com