SOLUTION: Under root(x^2+4*x-21) +under root(x^2-x-6)=under root (6*x^2 -5*x-39)

Algebra ->  Equations -> SOLUTION: Under root(x^2+4*x-21) +under root(x^2-x-6)=under root (6*x^2 -5*x-39)       Log On


   

Question 963997: Under root(x^2+4*x-21) +under root(x^2-x-6)=under root (6*x^2 -5*x-39)
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%28x%5E2%2B4x-21%29+%2Bsqrt%28x%5E2-x-6%29=sqrt%286x%5E2+-5x-39%29

Factor out the sqrt%28x-3%29 from all terms.
sqrt%28x%2B7%29+%2Bsqrt%28x%2B2%29=sqrt%286x%2B13%29
Square both sides,
%28x%2B7%29%2B2sqrt%28%28x%2B7%29%28x%2B2%29%29%2B%28x%2B2%29=6x%2B13
2x%2B9%2B2sqrt%28%28x%2B7%29%28x%2B2%29%29%2B%28x%2B2%29=6x%2B13
2sqrt%28%28x%2B7%29%28x%2B2%29%29=4x%2B4
2sqrt%28%28x%2B7%29%28x%2B2%29%29=4%28x%2B1%29
sqrt%28%28x%2B7%29%28x%2B2%29%29=2%28x%2B1%29
Square both sides,
%28x%2B7%29%28x%2B2%29=4%28x%5E2%2B2x%2B1%29
x%5E2%2B9x%2B14=4x%5E2%2B8x%2B4
3x%5E2%2B9x%2B14=4x%5E2%2B8x%2B4
3x%5E2-x-10=0
%28x-2%29%283x%2B5%29=0
Remember we factored out the sqrt%28x-3%29 before so that's still part of the solution.
We only removed it so that the calculation wouldn't be messier.
sqrt%28x-3%29%28x-2%29%283x%2B5%29=0
Three possible solutions:
sqrt%28x-3%29=0
x-3=0
x=3
Verifying,
sqrt%283%5E2%2B4%283%29-21%29+%2Bsqrt%283%5E2-3-6%29=sqrt%286%283%29%5E2+-5%283%29-39%29
sqrt%289%2B12-21%29+%2Bsqrt%289-9%29=sqrt%2854+-15-39%29
0%2B0=0
True, good solution.
.
.
x-2=0
x=2
Verifying,

sqrt%284%2B8+-21%29+%2Bsqrt%284-2-6%29=sqrt%28+24-5%282%29-39%29
sqrt%28-9%29+%2Bsqrt%28-4%29=sqrt%28+-25%29
Leads to a negative square root so not a real solution.
.
.
3x%2B5=0
3x=-5
x=+%28-5%2F3%29+
Verifying,

sqrt%2825%2F9-20%2F3-21%29+%2Bsqrt%2825%2F9%2B5%2F3-6%29=sqrt%2850%2F3%2B25%2F3+-39%29
sqrt%28-224%2F9%29+%2Bsqrt%28-14%2F9%29=sqrt%28-14%29
Leads to a negative square root so not a real solution.
.
.
Only one real solution, x=3