√ = square root, and yes the first part says the square root of a square root and the parenthesis indicate which part of the problem is in what square root.
Solve for x.
√[√(x + 5) + x] = 5
Square both sides.
√(x + 5) + x = 25
√(x + 5) = 25 - x
x + 5 = (25 - x)² = 625 - 50x + x²
0 = x² - 51x + 620
(x - 20)(x - 31) = 0
x = 20, 31
Plug back into the original equation to see if the solutions fit. We find x = 31 is an extraeous solution. It does not fit.
x = 20
0
.
ââx+5+x=5
ââ(x+x=5-5
ââ2x=0
ââx=0
ââ0
â0
0.
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Verified answer
Solve for x.
√[√(x + 5) + x] = 5
Square both sides.
√(x + 5) + x = 25
√(x + 5) = 25 - x
Square both sides.
x + 5 = (25 - x)² = 625 - 50x + x²
0 = x² - 51x + 620
(x - 20)(x - 31) = 0
x = 20, 31
Plug back into the original equation to see if the solutions fit. We find x = 31 is an extraeous solution. It does not fit.
x = 20
0
.
.
.
ââx+5+x=5
ââ(x+x=5-5
ââ2x=0
ââx=0
ââ0
â0
0.