Question

Solve the equation square root(x^2-5)-square root(x^2-8)=1

Accepted Answer

We have to solve sqrt(x^2 - 5) - sqrt(x^2 - 8) = 1
sqrt(x^2 - 5) - sqrt(x^2 - 8) = 1
square both the sides
x^2 - 5 + x^2 - 8 - 2* sqrt [(x^2 - 5)(x^2 - 8)]  = 1
=> 2x^2 - 14 - 2* sqrt [(x^2 - 5)(x^2 - 8)]  = 0
=> x^2 - 7 = sqrt [(x^2 - 5)(x^2 - 8)]
square both the sides
=> x^4 + 49 - 14x^2 = x^4 - 13x^2 + 40
=> 9 - x^2 = 0
=> x^2 = 9
=> x = 3 and x = -3
The required solutions are x = 3 and x = -3

Answer

We'll impose the constraints of existence of square roots:
x^2-5>=0
the expression is positive if x belongs to the ranges:
(-infinite ; -sqrt5]U[sqrt5 ; +infinite)
x^2-8>=0
the expression is positive if x belongs to the ranges:
(-infinite ; -sqrt8]U[sqrt8 ; +infinite)
The common intervals of admissible values for x are:
(-infinite ; -sqrt8]U[sqrt8 ; +infinite)
Now, we'll solve the equation. We'll move -sqrt(x^2-8) to the righ side:
sqrt(x^2-5) = sqrt(x^2-8) + 1
We'll raise to square both sides, to eliminate the square root from the left side:
x^2 - 5 = x^2 - 8 + 1 + 2sqrt(x^2-8)
We'll eliminate x^2 both sides:
7 - 5 = 2sqrt(x^2-8)
2 = 2sqrt(x^2-8)
We'll divide by 2:
sqrt(x^2-8) = 1
We'll raise to square again:
x^2 - 8 = 1
x^2 = 9
x1 = 3 and x2 = -3
Since both values belong to the intervals of admissible values, the solutions of the equation are: {-3 ; 3}.

algebra1

Solve the equation square root(x^2-5)-square root(x^2-8)=1

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