SOLUTION: Find the real solutions of the equation. (5x^2 - 6)^(1/4) = x I found x to be -sqrt{6} and sqrt{6}. The textbook answer is x = sqrt{2} and x = sqrt{3}. Help.

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Question 1207658: Find the real solutions of the equation.
(5x^2 - 6)^(1/4) = x
I found x to be -sqrt{6} and sqrt{6}.
The textbook answer is x = sqrt{2} and x = sqrt{3}.
Help.
Found 2 solutions by Theo, MathTherapy:
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
i solved it this way.

start with (5x^2 - 6) ^ (1/4) = x
raise both sides of the equation to the 4th power to get:
5x^2 - 6 = x^4
subtract the left side of the equation from both sides to get:
0 = x^4 - 5x^2 + 6
replace x^2 with k.
you get k^2 - 5k + 6 = 0
factor to get:
(k - 3) * (k - 2) = 0.
solve for k to get:
k = 3 or k = 2
replace k with x^2 to get:
x^2 = 3 or x^2 = 2
solve for x to get:
x = sqrt(3) or x = sqrt(2).

Answer by MathTherapy(10552) (Show Source):
You can put this solution on YOUR website!

Find the real solutions of the equation. (5x^2 - 6)^(1/4) = x I found x to be -sqrt{6} and sqrt{6}. The textbook answer is x = sqrt{2} and x = sqrt{3}. Help. Your answers are WRONG, but because you failed to show your work, no-one can determine where you made your mistake. Note that the left-side of the equation is being raised to a power (), so the right-side (x), CANNOT be negative (< 0). This will determine the correct answers. ---- Raising each side to the 4^th power 0 = (x² - 3)(x² - 2) As stated earlier, the right-side of the equation CANNOT be negative (< 0), so