How do i?!?

Let u = √x. Then x = u². This allows us to rewrite the equation as:

2u² - 2u - 40 = 0

Which is easily solvable by the quadratic formula.

u = (1/4)(2 ± √(4 + 320))

u = 1/2 ± (1/4)*18

u = 1/2 ± 9/2

u = 5 or -4

That means we have two solutions for x:

1) u = 5

√x = 5

x = 25

2) u = -4

√x = -4

This does not lead to a solution since square root is always positive. Thus, the only solution is x = 25.

2x - 2√x - 40 = 0

let a= √x

2a^2 -2a-40 = 0 --> a^2 - a - 20 = 0

a= 5 or a = -4

--> x = 25 or x = 16

2x - 40 = 2sqrt(x)

x - 20 = sqrt(x)

x^2 - 41x + 400 = 0

(x-25)(x-16) = 0

x=25 or x=16

However, x = 16 does not satisfy the original equation, so

x = 25 is the sole solution.

sub y = √x

y^2 = x

2y^2 - 2y - 40 = 0

y^2 - y - 20 = 0

(y - 5)(y + 4) = 0

y = 5 of y = -4 (but since y = √x, y cannot be negative)

So y = 5, which means x = 25 is the solution.

let u = √x

2u^2 - 2u - 40 = 0

u^2 - u - 20 = 0

(u-5)(u+4) = 0

u = 5 or -4

√x = 5 x = 25

√x = -4 x = 16