Question 75009
{{{2(x-5)^2 = 3}}}
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You can solve this by first dividing both sides by 2.  This removes the factor of 2 on the
left side. After the division by 2 the equation becomes:
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{{{(x-5)^2 = 3/2}}}
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Take the square root of both sides to get:
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{{{(x-5) = sqrt(3/2)}}}
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The right side of this equation generates both positive and negative answers.  So we end up
with two answers for x-5.  They are {{{x-5 = +sqrt(3/2)}}} and {{{x - 5 = -sqrt(3/2)}}}.
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Add +5 to both sides of both equations and you get:
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{{{x = 5 + sqrt(3/2)}}} and {{{x = 5 - sqrt(3/2)}}}
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One more thing.  It is against convention to leave a radical in the denominator.  So we need
to work on {{{sqrt(3/2)}}} because that is equal to {{{sqrt(3)/sqrt(2)}}}.  We can rationalize
this fraction by multiplying it by {{{sqrt(2)/sqrt(2)}}} which is the same as multiplying 
it by 1, so it doesn't change the value of the fraction.  This multiplication becomes:
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{{{(sqrt(3)/sqrt(2))*(sqrt(2)/sqrt(2)) = (sqrt(3)*sqrt(2))/(sqrt(2)*sqrt(2))}}}
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The denominator is rationalized to become 2 and the numerator becomes {{{sqrt(3*2)}}}
which becomes {{{sqrt(6)}}}.
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The result is {{{sqrt(6)/2}}} and when this is substituted into the equations for x, the
result is:
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{{{x = 5 + sqrt(6)/2}}} and {{{x = 5 - sqrt(6)/2}}}
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