Question

Question #51c56

Answer 1

`16x^4 + 232x^3 + 673x^2 - 818x + 41 = 0`

Explanation:

`sqrtx div sqrt(1-x)+sqrt(1-x)=5÷2`

`rArr sqrtx/sqrt(1 - x) + sqrt (1 - x) = 5/2`

Taking L.C.M

`rArr (sqrtx + 1 - x)/sqrt(1 - x) = 5/2`

Cross Multiply

`rArr 2 (sqrtx + 1 - x) = 5 sqrt(1 - x)`

`rArr 2sqrtx + 2 - 2x = 5 sqrt(1 - x)`

Collect Like Terms

`rArr 2sqrtx - 5 sqrt(1 - x) = 2x + 2`

Square both sides

`rArr (2sqrtx - 5 sqrt(1 - x))^2 = (2x + 2)^2`

`rArr (2sqrtx - 5 sqrt(1 - x)) (2sqrtx - 5 sqrt(1 - x)) = (2x + 2) (2x + 2)`

`rArr 4(x) - 20sqrt(1 - x) + 25(1 - x) = 4x^2 + 8x + 4`

`rArr 4(x) - 20sqrt(1 - x) + 25 - 25x = 4x^2 + 8x + 4`

`rArr- 20sqrt(1 - x) + 25 - 21x = 4x^2 + 8x + 4`

`rArr- 20sqrt(1 - x) = 4x^2 + 8x + 4 +21x - 25`

`rArr- 20sqrt(1 - x) = 4x^2 + 29x - 21`

Square both sides

`rArr (- 20sqrt(1 - x))^2 = (4x^2 + 29x - 21)^2`

`rArr 400 (1 - x) = (4x^2 + 29x - 21)^2`

`rArr 400 - 400x = (4x^2 + 29x - 21) (4x^2 + 29x - 21)`

`rArr 400 - 400x = 16x^4 + 232x^3 + 673x^2 - 1218x + 441`

Collect like terms

`rArr 16x^4 + 232x^3 + 673x^2 - 1218x + 400x + 441 - 400 = 0`

`rArr 16x^4 + 232x^3 + 673x^2 - 818x + 41 = 0`

Solve the polynomial above..

That's how far i could get, But in my own point of view, i strongly doubt the Authenticity of the question..

Answer 2

See below.

Explanation:

`sqrtx/sqrt(1-x)+sqrt(1-x)=2.5` Calling `y = 1-x` we have

`sqrt(1-y)+y=2.5 sqrty` or

`1-y=(2.5 sqrty-y)^2` or

`1-y=2.5^2y-5y sqrty+y^2` and now making `xi=sqrty` we have

`xi^4-5xi^3+(1+2.5^2)xi^2-1=0` This polynomial has two real roots

`xi = {-0.332869,0.43602}` giving

`y = {0.110802,0.190114}` and consequently

`x = {0.889198,0.809886}`

but substituting into the original equation only `x = 0.809886` is feasible, so the solution is

`x = 0.809886`