SOLUTION: solve for x {{{ x^2-6x-160=0 }}}

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Question 118906: solve for x +x%5E2-6x-160=0+
Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Given to solve:
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+x%5E2-6x-160=0+
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This equation can be solved by factoring. Look for factors of -160 that can be combined to
give -6. If you think about this -160 can be factored to -16 and + 10. If these two numbers
are added the result is -6. Using this information you can factor the left side of the equation
to get:
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%28x+-+16%29%2A%28x+%2B+10%29+=+0
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If you multiply out the two factors on left side you will get the given equation ... which is
as it should be.
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If either of the factors is equal to zero, the equation will be true ... because a multiplication
by zero on the left side will make the entire left side equal to zero and therefore, the
left side will equal the zero on the right side.
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So let's set each of the factors equal to zero and this will make the equation true.
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First:
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x+-+16+=+0
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Add 16 to both sides of this equation and you get for x:
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x+=+16
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Next set the other factor equal to zero:
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+x+%2B+10+=+0
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Subtract 10 from both sides and you get:
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x+=+-10
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So you have two solutions for x ... x = 16 and x = -10. You can check these two answers by
returning to the original (given) equation and substituting each of these answers for
x to see if they make the left side of the equation become zero and therefore equal to the
right side.
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The given equation is x%5E2+-+6x+-+160+=+0
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Letting x be 16 makes the left side:
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16%5E2+-+6%2816%29+-+160+=+256+-+96+-+160+=+256+-+256+=+0
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Next let x be -10 makes the left side:
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%28-10%29%5E2+-6%28-10%29+-+160+=+100+%2B+60+-+160+=+160+-+160+=+0
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So for either value of x (16 and -10) the left side of the given equation becomes zero
and therefore equal to the right side of the equation ... making the equation true.
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This means that the two answers we found are correct.
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Hope this helps you to understand this problem and one way that it can be solved (factoring).
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