Solve x-4=sqrt{x+2}

$$x-4 = \sqrt{ x+2 }$$

Solve for x

$$x=7$$

Solution Steps

Square both sides of the equation.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$x^{2}-8x+16=\left(\sqrt{x+2}\right)^{2}$$

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

$$x^{2}-8x+16=x+2$$

Subtract x from both sides.

$$x^{2}-8x+16-x=2$$

Combine -8x and -x to get -9x.

$$x^{2}-9x+16=2$$

Subtract 2 from both sides.

$$x^{2}-9x+16-2=0$$

Subtract 2 from 16 to get 14.

$$x^{2}-9x+14=0$$

To solve the equation, factor x^{2}-9x+14 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.

$$a+b=-9 ab=14$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 14.

$$-1,-14 -2,-7$$

Calculate the sum for each pair.

$$-1-14=-15 -2-7=-9$$

The solution is the pair that gives sum -9.

$$a=-7 b=-2$$

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

$$\left(x-7\right)\left(x-2\right)$$

To find equation solutions, solve x-7=0 and x-2=0.

$$x=7 x=2$$

Substitute 7 for x in the equation x-4=\sqrt{x+2}.

$$7-4=\sqrt{7+2}$$

Simplify. The value x=7 satisfies the equation.

$$3=3$$

Substitute 2 for x in the equation x-4=\sqrt{x+2}.

$$2-4=\sqrt{2+2}$$

Simplify. The value x=2 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-2=2$$

Equation x-4=\sqrt{x+2} has a unique solution.

$$x=7$$