how would you solve for x in the following?

Question

sqrt(1-2x)+sqrt(x+5)=2

Answer 1

sqrt(1-2x)+ sqrt(x+5) = 2.
hence, sqrt [ (1-2x)*(x+5)] = 2.
Sqrt (-2x^2 -9x+5)= 2
Now after squaring we get,
-2x^2 - 9x+ 5= 4
2x^2+ 9x -5= -4
2x^2 + 9x -5+4=0
2x^2 +9x-1=0
Using quadratic formula,
x=-b+-sqrt b^2 -4ac/2a
In the above equation, a=2, b=9, c= -1.
The above equation will have two solutions,
Hence, x= -9+sqrt89/4 or x = -9 - sqrt89/4.
hope it helps,
nt.

Answer 2

sqrt(1-2x)+sqrt(x+5)=2
when you square both sides you get,
( 1- 2x ) + 2 sqrt ( 1- 2x) ( x+5 ) + ( x+5) = 4
6 -x + 2 sqrt ( 1- 2x) ( x+5 ) + ( x+5)=4
leave the square root in the left hand side and move all other terms to the right hand side
2 sqrt ( 1- 2x) ( x+5 ) + ( x+5) = 4 - 6 +x
2 sqrt ( 1- 2x) ( x+5 ) + ( x+5) = (-2 +x)
Now, square both sides again,
( 1-2x ) ( x +5) = ( -2 +x )^2
x + 5 - 2x^2 - 10 = 4 - 4x + x^2
-2x^2 - x^2 +x +4x +5 - 10 -4= 0
- 3x^2 +5x - 14 = 0
3x^2 -5x + 14 = 0
this equation has no real solution,it has to imaginary solutions,
0.83 + 1.99 i and 0.83 - 1.99 i

Answer 3

√(1 - 2x) + √(x + 5) = 2
√(x + 5) = 2 - √(1 - 2x)
Square both sides:
x + 5 = 4 - 4√(1 - 2x) + 1 - 2x
Simplify:
3x = - 4√(1 - 2x)
Square both sides:
9x² = 16(1 - 2x)
9x² + 32x - 16 = 0

x = (-32 ± √[32² - 4(9)(-16)] ) /18 = (-32±40)/18 = 4/9 or -4

However, if you substitute each of these into the original equation, you will find that they do NOT solve that equation. The problem arises from squaring both sides of an equation (in this case, we had to do it twice). This has the potential for introducing spurious solutions (essentially because when you square a negative quantity, you get the same result as if the original quantity had been positive, or vice versa). So this problem has no solution.

Answer 4

square all sides to get rid of the roots:

(1-2x) + (x+5) = 4

1-2x + x +5 = 4

-x +2 = 0

2 = x

Answer 5

sqrt(1-2x)+sqrt(x+5)=2

square both sides
[sqrt(1-2x)+sqrt(x+5)]^2=(2)^2
[sqrt(1-2x)]^2+[sqrt(x+5)]^2+[2*sqrt(x+5)sqrt(1-2x)]=4
[1-2x]+[x+5]+[2*sqrt{(x+5)(1-2x)}]=4
[1-2x+x+5]+[2*sqrt{(x+5-2x^2-10x)}]=4
[6-x]+[2*sqrt{(-2x^2-9x+5)}]=4
-x+[2*sqrt{(-2x^2-9x+5)}]=4-6
-x+[2*sqrt{(-2x^2-9x+5)}]=-2
[2*sqrt{(-2x^2-9x+5)}]=x-2
2sqrt{(-2x^2-9x+5)}]=x-2

square both sides again
[2sqrt{(-2x^2-9x+5)}]^2=(x-2)^2
4[sqrt{(-2x^2-9x+5)}]^2=x^2+4-4x
4(-2x^2-9x+5)=x^2+4-4x
-8x^2-36x+20=x^2+4-4x
-8x^2-36x+20-x^2-4+4x=0
-9x^2-32x+16=0
-9x^2-36x+4x+16=0
-9x(x+4)+4(x+4)=0
(-9x+4)(x+4)=0
so
(-9x+4)=0
-9x+4=0
-9x=-4
9x=4
x=4/9

or

x+4=0
x=-4

so x =4/9 or -4