SOLUTION: Solve each equation for x. 2(x-5)^2 = 3 My anwer was:x={{{2sqrt(6+10)/2}}},x={{{2sqrt(-6+10)/2}}}. Is this correct? Thanks.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Solve each equation for x. 2(x-5)^2 = 3 My anwer was:x={{{2sqrt(6+10)/2}}},x={{{2sqrt(-6+10)/2}}}. Is this correct? Thanks.       Log On


   

Question 75009This question is from textbook Beginning Algebra
: Solve each equation for x.
2(x-5)^2 = 3
My anwer was:x=2sqrt%286%2B10%29%2F2,x=2sqrt%28-6%2B10%29%2F2. Is this correct? Thanks. This question is from textbook Beginning Algebra

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
2%28x-5%29%5E2+=+3
.
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:
.
%28x-5%29%5E2+=+3%2F2
.
Take the square root of both sides to get:
.
%28x-5%29+=+sqrt%283%2F2%29
.
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+=+%2Bsqrt%283%2F2%29 and x+-+5+=+-sqrt%283%2F2%29.
.
Add +5 to both sides of both equations and you get:
.
x+=+5+%2B+sqrt%283%2F2%29 and x+=+5+-+sqrt%283%2F2%29
.
One more thing. It is against convention to leave a radical in the denominator. So we need
to work on sqrt%283%2F2%29 because that is equal to sqrt%283%29%2Fsqrt%282%29. We can rationalize
this fraction by multiplying it by sqrt%282%29%2Fsqrt%282%29 which is the same as multiplying
it by 1, so it doesn't change the value of the fraction. This multiplication becomes:
.

.
The denominator is rationalized to become 2 and the numerator becomes sqrt%283%2A2%29
which becomes sqrt%286%29.
.
The result is sqrt%286%29%2F2 and when this is substituted into the equations for x, the
result is:
.
x+=+5+%2B+sqrt%286%29%2F2 and x+=+5+-+sqrt%286%29%2F2
.