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

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You need raise to square both sides such that:

You need to expand the square to the right such that:

You need to eliminate both sides such that:

You should keep the term that contains x to the right side and you need to move the constant terms to the left side such that:

You need to remember that the value of x is good only if it makes the radicand positive such that:

**Hence, the solution to the equation is .**

sqrt (x^2 - 3) = x-5

First let us square both sides:

[sqrt(x^2-3)]^2 = (x-5)^2

==> (x^2 -3) = x^2 - 10x + 25

Reduce similar :

==> -3 = -10x + 25

==> 10x = 28

Now divide by 10:

==> x = 28/10 = 14/5

==> x= 14/5 = 2.8

For the beginning, we'll impose constraints of existence of square root:

x^2-3 > 0

x^2-3 = 0

x^2 = 3

x1 = +sqrt3

x2 = -sqrt3

The expression x^2-3 is positive if x is in the intervals;

(-inf. , -sqrt3) U (sqrt3, +inf.)

Now, we'll solve the equation. We'll raise to square both sides, to eliminate the square root from the left side:

[sqrt (x^2-3)]^2 = (x-5)^2

x^2 - 3 = x^2 - 10x + 25

We'll eliminate x^2 from both sides:

-3 = -10x + 25

We'll add 3 both sides:

-10x + 25 + 3 = 0

-10x + 28 = 0

-10x = -28

x = 28/10

x = 2.8

**Because 2.8 > sqrt3, the solution belongs to the interval of admissible values of x, so the solution is valid.**

sqrt(x^2-3) = x-5. To find x.

Silution:

sqrt(x^2-3) = (x-5)

square both sides

x^2-3 = (x-5)^2

(x-5)^2- (x^2-3) = 0

x^2-10x+25-x^2+3 =

-10x +28 = 0

-10x = -28

x = 28/10 = 2.8

Given:

sqrt(x^2 - 3) = x - 5

Taking squares of both sides of this equation:

x^2 - 3 = (x - 5)^2

==>x^2 - 3 = x^2 -10x + 25

Transferring all the terms containing x on the left hand side, and other terms on the right hand side:

==> x^2 - x^2 + 10x = 25 + 3

==>10x = 28

Dividing both sides by 10:

x = 28/10 = 2.8