How do i?!?

Question

How do I solve this equation by making an appropriate substitution?!

2x - 2√x - 40 = 0

Answer 1

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.

Answer 2

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

Answer 3

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.

Answer 4

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.

Answer 5

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