Solve sqrt{16-3x}=x-2

$$\sqrt { 16 - 3 x } = x - 2$$

Solve for x

$$x=4$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{16-3x}\right)^{2}=\left(x-2\right)^{2}$$

Calculate \sqrt{16-3x} to the power of 2 and get 16-3x.

$$16-3x=\left(x-2\right)^{2}$$

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

$$16-3x=x^{2}-4x+4$$

Subtract x^{2} from both sides.

$$16-3x-x^{2}=-4x+4$$

Add 4x to both sides.

$$16-3x-x^{2}+4x=4$$

Combine -3x and 4x to get x.

$$16+x-x^{2}=4$$

Subtract 4 from both sides.

$$16+x-x^{2}-4=0$$

Subtract 4 from 16 to get 12.

$$12+x-x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}+x+12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

$$a+b=1 ab=-12=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 1.

$$a=4 b=-3$$

Rewrite -x^{2}+x+12 as \left(-x^{2}+4x\right)+\left(-3x+12\right).

$$\left(-x^{2}+4x\right)+\left(-3x+12\right)$$

Factor out -x in the first and -3 in the second group.

$$-x\left(x-4\right)-3\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(-x-3\right)$$

To find equation solutions, solve x-4=0 and -x-3=0.

$$x=4 x=-3$$

Substitute 4 for x in the equation \sqrt{16-3x}=x-2.

$$\sqrt{16-3\times 4}=4-2$$

Simplify. The value x=4 satisfies the equation.

$$2=2$$

Substitute -3 for x in the equation \sqrt{16-3x}=x-2.

$$\sqrt{16-3\left(-3\right)}=-3-2$$

Simplify. The value x=-3 does not satisfy the equation because the left and the right hand side have opposite signs.

$$5=-5$$

Equation \sqrt{16-3x}=x-2 has a unique solution.

$$x=4$$