Looking at t³ - 5t - 2 = 0 we see t = -2 is a solution so we can write it as (t + 2) (t² - 2t - 1) = 0. But t needs to be non-negative for x to be real so there is only one solution for t: √x = t = 1 + √2 and x = t² = 3 + 2√2 and finally x - 2√x = 3 + 2√2 - 2 - 2√2 = 1.
Sorry, I don't get why √x ≠ -2. If x is negative then √x will be a complex number, but the converse isn't necessarily true. for √x = -2, x = 4. Shouldn't (√x)² - 2√x = (-2)² - 2(-2) = 8, also be a valid solution?
I will try the following : y=√x and E = x-2√x = y^2-2y => y^2-2/y=5 y^3-2y^2+2y^2-5y-2=0 y*E+2y^2-4y+4y-5y-2=0 y*E+2E -y -2=0 y(E-1)+2(E-1)=0 (y+2)(E-1)=0 so E must be 1
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Answer = 1
I used substitution because substitution is awesome!
Looking at t³ - 5t - 2 = 0 we see t = -2 is a solution so we can write it as (t + 2) (t² - 2t - 1) = 0. But t needs to be non-negative for x to be real so there is only one solution for t: √x = t = 1 + √2 and x = t² = 3 + 2√2 and finally x - 2√x = 3 + 2√2 - 2 - 2√2 = 1.
Sorry, I don't get why √x ≠ -2. If x is negative then √x will be a complex number, but the converse isn't necessarily true. for √x = -2, x = 4. Shouldn't (√x)² - 2√x = (-2)² - 2(-2) = 8, also be a valid solution?
The square root of a positive number is positive by definition so it can't be -2.
4-(-1)= 5
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I will try the following :
y=√x and E = x-2√x = y^2-2y => y^2-2/y=5 y^3-2y^2+2y^2-5y-2=0
y*E+2y^2-4y+4y-5y-2=0
y*E+2E -y -2=0
y(E-1)+2(E-1)=0
(y+2)(E-1)=0
so E must be 1
1
X= ?