# What are the real solutions of square root(x^2+9)=x-2 ?

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We have to solve sqrt(x^2 + 9) = x - 2

sqrt(x^2 + 9) = x - 2

square both the sides

x^2 + 9 = (x - 2)^2

=> x^2 + 9 = x^2 + 4 - 4x

=> x^2 - x^2 + 4x = 4 - 9

=> 4x = -5

=> x = -5/4

**The solution of the equation is x = -5/4**

Since the radicand (x^2 + 9) is positive for any real value of x, we don't have to impose the constraint of existence of the square root.

We'll raise to square both sides, to eliminate the square root:

x^2 + 9 = (x-2)^2

We'll expand the square from the right, using the formula:

(a-b)^2 = a^2 - 2ab + b^2

(x-2)^2 = x^2 - 4x + 4

The equation will become:

x^2 + 9 = x^2 - 4x + 4

We'll subtract both sides x^2 - 4x + 4:

x^2 + 9 - x^2 + 4x - 4 = 0

We'll eliminate like terms:

4x + 5 = 0

We'll shift 5 to the right:

4x = -5

x = -5/4

**The equation has only a real solution and this is x = -5/4.**