Solve sqrt{40-3x}=x-4

$$\sqrt{ 40-3x } = x-4$$

Solve for x

$$x=8$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{40-3x}\right)^{2}=\left(x-4\right)^{2}$$

Calculate \sqrt{40-3x} to the power of 2 and get 40-3x.

$$40-3x=\left(x-4\right)^{2}$$

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

$$40-3x=x^{2}-8x+16$$

Subtract x^{2} from both sides.

$$40-3x-x^{2}=-8x+16$$

Add 8x to both sides.

$$40-3x-x^{2}+8x=16$$

Combine -3x and 8x to get 5x.

$$40+5x-x^{2}=16$$

Subtract 16 from both sides.

$$40+5x-x^{2}-16=0$$

Subtract 16 from 40 to get 24.

$$24+5x-x^{2}=0$$

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

$$-x^{2}+5x+24=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+24. To find a and b, set up a system to be solved.

$$a+b=5 ab=-24=-24$$

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 -24.

$$-1,24 -2,12 -3,8 -4,6$$

Calculate the sum for each pair.

$$-1+24=23 -2+12=10 -3+8=5 -4+6=2$$

The solution is the pair that gives sum 5.

$$a=8 b=-3$$

Rewrite -x^{2}+5x+24 as \left(-x^{2}+8x\right)+\left(-3x+24\right).

$$\left(-x^{2}+8x\right)+\left(-3x+24\right)$$

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

$$-x\left(x-8\right)-3\left(x-8\right)$$

Factor out common term x-8 by using distributive property.

$$\left(x-8\right)\left(-x-3\right)$$

To find equation solutions, solve x-8=0 and -x-3=0.

$$x=8 x=-3$$

Substitute 8 for x in the equation \sqrt{40-3x}=x-4.

$$\sqrt{40-3\times 8}=8-4$$

Simplify. The value x=8 satisfies the equation.

$$4=4$$

Substitute -3 for x in the equation \sqrt{40-3x}=x-4.

$$\sqrt{40-3\left(-3\right)}=-3-4$$

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

$$7=-7$$

Equation \sqrt{40-3x}=x-4 has a unique solution.

$$x=8$$