Solve x-2=sqrt{x+10}

$$x-2= \sqrt{ x+10 }$$

Solve for x

$$x=6$$

Solution Steps

Square both sides of the equation.

$$\left(x-2\right)^{2}=\left(\sqrt{x+10}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

$$x^{2}-4x+4=\left(\sqrt{x+10}\right)^{2}$$

Calculate \sqrt{x+10} to the power of 2 and get x+10.

$$x^{2}-4x+4=x+10$$

Subtract x from both sides.

$$x^{2}-4x+4-x=10$$

Combine -4x and -x to get -5x.

$$x^{2}-5x+4=10$$

Subtract 10 from both sides.

$$x^{2}-5x+4-10=0$$

Subtract 10 from 4 to get -6.

$$x^{2}-5x-6=0$$

To solve the equation, factor x^{2}-5x-6 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-5 ab=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -5.

$$a=-6 b=1$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-6\right)\left(x+1\right)$$

To find equation solutions, solve x-6=0 and x+1=0.

$$x=6 x=-1$$

Substitute 6 for x in the equation x-2=\sqrt{x+10}.

$$6-2=\sqrt{6+10}$$

Simplify. The value x=6 satisfies the equation.

$$4=4$$

Substitute -1 for x in the equation x-2=\sqrt{x+10}.

$$-1-2=\sqrt{-1+10}$$

Simplify. The value x=-1 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-3=3$$

Equation x-2=\sqrt{x+10} has a unique solution.

$$x=6$$