SOLUTION: Can you please help solve for the following: {{{matrix(1,3,sqrt(14 - x), "=", 6 - sqrt(x + 4))}}}

Algebra ->  Square-cubic-other-roots -> SOLUTION: Can you please help solve for the following: {{{matrix(1,3,sqrt(14 - x), "=", 6 - sqrt(x + 4))}}}      Log On


   

Question 179467: Can you please help solve for the following:
matrix%281%2C3%2Csqrt%2814+-+x%29%2C+%22=%22%2C+6+-+sqrt%28x+%2B+4%29%29
Answer by Edwin McCravy(20086) About Me  (Show Source):
You can put this solution on YOUR website!

matrix%281%2C3%2Csqrt%2814+-+x%29%2C+%22=%22%2C+6+-+sqrt%28x+%2B+4%29%29
A radical term is already isolated on the left, so square both sides:

Take away the radical and exponent on the left, write the square on
the right as the product of the base times itself:

Use "FOIL" on the right:

Combine like terms on the right, take away radical and exponent
in the last term on the right:
matrix%281%2C3%2C14-x%2C+%22=%22%2C+36-12sqrt%28x%2B4%29%2Bx%2B4+++%29
Isolate the radical term on the left side:
matrix%281%2C3%2C12sqrt%28x%2B4%29%2C%22=%22%2C36%2Bx%2B4-14%2Bx+%29
Combine like terms:
matrix%281%2C3%2C12sqrt%28x%2B4%29%2C%22=%22%2C26%2B2x+%29
Notice that every term can be divided by 2, so do so:
matrix%281%2C3%2C6sqrt%28x%2B4%29%2C%22=%22%2C13%2Bx+%29
Square both sides:
matrix%281%2C3%2C%286sqrt%28x%2B4%29%29%5E2%2C%22=%22%2C%2813%2Bx%29%5E2+%29
matrix%281%2C3%2C6%5E2%28sqrt%28x%2B4%29%5E2%29%2C%22=%22%2C%2813%2Bx%29%2813%2Bx%29+%29
matrix%281%2C3%2C36%28x%2B4%29%2C%22=%22%2C13%5E2%2B13x%2B13x%2Bx%5E2+%29
matrix%281%2C3%2C36x%2B144%2C%22=%22%2C169%2B26x%2Bx%5E2%29
Get 0 on the left:
matrix%281%2C3%2C0%2C%22=%22%2C169%2B26x%2Bx%5E2-36x-144%29
matrix%281%2C3%2C0%2C%22=%22%2C25-10x%2Bx%5E2%29
Rearrange right side in descending order:
matrix%281%2C3%2C0%2C%22=%22%2Cx%5E2-10x%2B25%29
Switch right side and left side:
matrix%281%2C3%2Cx%5E2-10x%2B25%2C%22=%22%2C0%29
Factor left side:
matrix%281%2C3%2C%28x-5%29%28x-5%29%2C%22=%22%2C0%29
Use zero-factor principle:
matrix%282%2C3%2C%0D%0Ax-5=0%2C%22++++++%22%2Cx-5=0%2C%0D%0A++x=5%2C%22++++++%22%2Cx=5%29   

We get one solution x=5
However this might be an extraneous
(bogus) solution, for sometimes you
get those when you square both sides
of an equation.  So we must check
it to see if it satisfies the original
radical equation:

matrix%281%2C3%2Csqrt%2814+-+x%29%2C+%22=%22%2C+6+-+sqrt%28x+%2B+4%29%29
matrix%281%2C3%2Csqrt%2814+-+5%29%2C+%22=%22%2C+6+-+sqrt%285+%2B+4%29%29
matrix%281%2C3%2Csqrt%289%29%2C+%22=%22%2C+6+-+sqrt%289%29%29
matrix%281%2C3%2C3%2C+%22=%22%2C+6+-+3%29
matrix%281%2C3%2C3%2C+%22=%22%2C+3%29 

It does, so x=5 is the only solution.

Edwin