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

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### 2 Answers

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**

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}.**