Asked by Jonathan
What is the real solutions of this equation?
sqrt(2x+3)-sqrt(x+2)=2
sqrt(2x+3)-sqrt(x+2)=2
Answers
Answered by
drwls
First square both sides
2x+3 + x+2 -2sqrt([2x+3)(x+2)] = 4
3x + 5 -2sqrt[(2x+3)(x+2)] = 4
2sqrt[(2x+3)(x+2)] = 3x +1
Now square both sides again to get rid of the radical, and solve the resulting quadratic equation
4(2x+3)(x+2) = 9x^2 + 6x + 1
8x^2 + 28x + 24 = 9x^2 +6x = 1
x^2 -22x -23 = 0
(x-23)(x+1) = 0
x = 23 and x=-1 are the solutions. Note that for the -1 solution, you must take a negative square root of +1 to satisfy the original equation..
2x+3 + x+2 -2sqrt([2x+3)(x+2)] = 4
3x + 5 -2sqrt[(2x+3)(x+2)] = 4
2sqrt[(2x+3)(x+2)] = 3x +1
Now square both sides again to get rid of the radical, and solve the resulting quadratic equation
4(2x+3)(x+2) = 9x^2 + 6x + 1
8x^2 + 28x + 24 = 9x^2 +6x = 1
x^2 -22x -23 = 0
(x-23)(x+1) = 0
x = 23 and x=-1 are the solutions. Note that for the -1 solution, you must take a negative square root of +1 to satisfy the original equation..
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