Solve sqrt{40-3x}=x

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

Solve for x

$$x=5$$

Solution Steps

Square both sides of the equation.

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

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

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

Subtract x^{2} from both sides.

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

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 negative, the negative number has greater absolute value than the positive. 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.

$$a=5 b=-8$$

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

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

Factor out x in the first and 8 in the second group.

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

Factor out common term -x+5 by using distributive property.

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

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

$$x=5 x=-8$$

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

$$\sqrt{40-3\times 5}=5$$

Simplify. The value x=5 satisfies the equation.

$$5=5$$

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

$$\sqrt{40-3\left(-8\right)}=-8$$

Simplify. The value x=-8 does not satisfy the equation because the left and the right hand side have opposite signs.

$$8=-8$$

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

$$x=5$$