Question

Solve for x sqrt(2x+5) = sqrt(x+2) + sqrt(2x-3).

Accepted Answer

sqrt(2x+5) = sqrt(x+2) + sqrt(2x-3)
Let us square both sides:
==> (2x+5) = (x+2) + 2sqrt(x+2)(2x-3) + (2x-3)
==> (2x+5) = x+2 + 2x -3 + 2sqrt(2x^2 +x -6)
==> (2x+5) = 3x -1 + 2sqrt(2x^2 + x -6)
==> -x +6 = 2sqrt(x+2)(2x-3)
Square both sides again:
==> x^2 -12x + 36 = 4*(2x^2 + x - 6)
==> x^2 -12x +36 = 8x^2 +4x - 24
Group similar terms:
==> 7x^2 +16x -60 = 0
==> x1= [-16 + sqrt(25d + 4*7*60)]/14= (-16 + 44)14
             = 28/14= 2
==> x2= (-16 -44)/14= -60/14= -30/7
==> the solutions are:
x1= 2     and x2= -30/7

Answer

sqrt(2x+5) = sqrt(x+2)+sqrt(2x-3). To solve for x.
We  squaring both sides we get
2x+5 = x+2 + 2sqrt{(x+2)(2x-3)} +2x-3
2x+5  = 3x-1 +2sqrt{x+2)(2x-3)}
2x+5-3x+1 = 2sqrt{(x+2)(2x-3)}
-x+6 = 2sqrt{x+2)(2x-3)}. Again sqre both sides:
(-x+6)^2 = 4(x+2)(2x-3)
x^2-24x+36= 4{2x^2 -3x+4x-6}
x^2-24x+36 = 8x^2 +4x-24
0 = (8-1)x^2 +(24+4)x -24-36
7x^2 +28x-60 = 0
x =  {-28 +or- sqrt(28^2 - - 4*7*60}/2*7
x1 = {-28 + sqrt(2864)}/14.
x2 {-28 -sqrt(2864)}/14

Answer

Before solving the equation, we'll set the constraint of existence of the sqrt.
2x + 5 >=0
2x>=-5
x>=-5/2
x+2>=0
x>=-2
2x-3>=0
2x>=3
x>=3/2
The interval of admissible values for x is [3/2 , +inf.)
Now, we'll solve the equation, raising to square both sides, in order to eliminate the square root:
[sqrt(2x+5)]^2 = [sqrt(x+2) + sqrt(2x-3)]^2
2x + 5 = x + 2 + 2x - 3 + 2sqrt(x+2)(2x-3)
We'll keep only the sqrt to the right side, the rest of the terms we'll move them to the left side:
2x + 5 -x - 2 - 2x + 3 = 2sqrt(x+2)(2x-3)
We'll eliminate like terms:
6-x = 2sqrt(x+2)(2x-3)
We'll raise to square again:
(6-x)^2 = 4*(x+2)(2x-3)
We'll expand the square and we'll remove the brackets:
36 - 12x + x^2 = 8x^2 + 4x - 24
We'll move all terms to one side:
36 - 12x + x^2 - 8x^2 - 4x + 24 = 0
We'll combine like terms:
-7x^2 - 16x + 60 = 0
We'll multiply by -1:
7x^2 + 16x - 60 = 0
We'll apply the quadratic formula:
x1 = (-16 + 44) / 14
x1 = 2
x2 = (-16 - 44) / 14
x2 = -60 / 14
x2 = -30/7
Since the value of the first root belongs to the interval of admissible values of x, the equation will have only one solution, namely x = 2.

Math

Solve for x sqrt(2x+5) = sqrt(x+2) + sqrt(2x-3).

Expert Answers

We’ll help your grades soar

Related Questions

x^3+2x^2-x-2=0 solve for x

sqrt(x^2+5) = 2x-3 solve for x

lim x->infinity ((sqrt(x^2+5))-(sqrt(x^2+3)))

Solve for X. e^2x - 3e^x + 2=0

Solve x.e for sqrt (x^2-3) = x-5.Solve x.e for sqrt (x^2-3) = x-5.