Question 118906
Given to solve:
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{{{ x^2-6x-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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{{{(x - 16)*(x + 10) = 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 + 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^2 - 6x - 160 = 0}}}
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Letting x be 16 makes the left side:
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{{{16^2 - 6(16) - 160 = 256 - 96 - 160 = 256 - 256 = 0}}}
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Next let x be -10 makes the left side:
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{{{(-10)^2 -6(-10) - 160 = 100 + 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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