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

[@wryanihad]

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

[@yakupbuyankara5903]

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*

[@davidshen5916]

X^2-X=17（X+Sqrt（X））， （X+Sqrt（X））（X-Sqrt（X））=17（X+Sqrt（X））， （X+Sqrt（X））（X-Sqrt（X）-17）=0， X+Sqrt（X）=0， X=0， X- Sqrt（X）=0， X-Sqrt（X）-17=0， X-Sqrt（X）=17