Working With A Nice Radical Expression

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[@danieldepaula6930]

Another way would be just calling y = sqrt(x). This gives us the equation y⁴-18y²-17y=0. By inspection, we quickly notice that y=-1 is a possibility. So, by Briot-Ruffini, we have y³-y²-17y=0, or y(y²-y-17)=0, and therefore, y=0 or y=(1±sqrt(69))/2. Returning to x, as we have y = sqrt(x), the only admissible values are those who are ≥0. So sqrt(x) can be 0 (and then, x=0, and therefore, x-sqrt(x)=0) or (1+sqrt(69))/2 (and then, x=(35+sqrt(69))/2, and therefore, x-sqrt(x)=17).

[@mcwulf25]

Method 2 slightly quicker if you factor out sqrt(x)+1 sooner.

[@alexandermorozov2248]

Zero is asnwer too.

[@VictorPensioner]

x² - 18x = 17√x => (x² - x) - (17x + 17√x) = 0
a² - b² = (a-b)(a+b)
If
a = x,
b = √x
then
(x² - x) = (x - √x)(x + √x)
and
(x - √x)(x + √x) - 17(x + √x) = 0
or
(x + √x)*(x - √x - 17) = 0
Therefore
1) x + √x = 0 => x = 0 and x - √x = 0
2) x - √x = 17

[@user-wj1qb3qu1y]

An edit on second method
x²-x=17(x+sqrx)
(x-sqrx)(x+sqrx)=17(x+sqrx)
(x-sqrx)=17

[@misterdubity3073]

Let A=x-√x; A*(x+√x) = x^2 - x; Eqn: x^2 - x = 17(x+√x); A*(x+√x) =17( 17(x+√x); A = 17

[@SidneiMV]

x² - 18x = 17√x
find E = x - √x
√x⁴ - 18√x² = 17√x
√x(√x³ - 18√x - 17) = 0
√x = 0 => x = 0 => *E = 0*
√x³ - 18√x - 17 = 0
√x³ + 1 - 18√x - 18 = 0
(√x + 1)(√x² - √x + 1) - 18(√x + 1) = 0
√x² - √x - 17 = 0 => *E = x - √x = 17*

[@yakupbuyankara5903]

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