Solve sqrt{3x+40}=x

$$\sqrt { 3 x + 40 } = x$$

Solve for x

$$x=8$$

Solution Steps

Square both sides of the equation.

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

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

$$3x+40=x^{2}$$

Subtract x^{2} from both sides.

$$3x+40-x^{2}=0$$

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

$$-x^{2}+3x+40=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+40. To find a and b, set up a system to be solved.

$$a+b=3 ab=-40=-40$$

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 -40.

$$-1,40 -2,20 -4,10 -5,8$$

Calculate the sum for each pair.

$$-1+40=39 -2+20=18 -4+10=6 -5+8=3$$

The solution is the pair that gives sum 3.

$$a=8 b=-5$$

Rewrite -x^{2}+3x+40 as \left(-x^{2}+8x\right)+\left(-5x+40\right).

$$\left(-x^{2}+8x\right)+\left(-5x+40\right)$$

Factor out -x in the first and -5 in the second group.

$$-x\left(x-8\right)-5\left(x-8\right)$$

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

$$\left(x-8\right)\left(-x-5\right)$$

To find equation solutions, solve x-8=0 and -x-5=0.

$$x=8 x=-5$$

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

$$\sqrt{3\times 8+40}=8$$

Simplify. The value x=8 satisfies the equation.

$$8=8$$

Substitute -5 for x in the equation \sqrt{3x+40}=x.

$$\sqrt{3\left(-5\right)+40}=-5$$

Simplify. The value x=-5 does not satisfy the equation because the left and the right hand side have opposite signs.

$$5=-5$$

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

$$x=8$$