sqrt(3-x) + sqrt (3+x) = x
(sqrt(3-x) + sqrt (3+x))^2 = x^2 but check at the end because -something ^2 is the same as something^2
(3-x) +2 sqrt(3-x)sqrt(3+x) +(3+x) = x^2
6 + 2 sqrt[(3-x)(3+x)] = x^2
6 + 2 sqrt[ 9 -x^2] = x^2
2 sqrt [ 9-x^2] = x^2-6
4 (9-x^2) = x^4 - 12 x^2 + 36
36 -4 x^2 = x^4 -12 x^2 + 36
x^4 - 8 x^2 = 0
x^2 (x^2-8) = 0
x = 0 or 0 or 2 sqrt 2 or -2 sqrt 2
try 0, no
2 sqrt 2 = 2.82 = yes
-2sqrt 2 = -2.82 yes
those both work but 0 does not
can you show me the steps how to do this?
square root of 3-x + square root of 3+x = x
2 answers
Square both sides. Then bring all terms except the remaining square root to one side and then square again. Put x^2 = t in that equation and solve the quadratic equation in t.
3 answers