Solve x=sqrt{-3x+40}

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

Solve for x

$$x=5$$

Solution Steps

Square both sides of the equation.

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

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

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

Add 3x to both sides.

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

Subtract 40 from both sides.

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

To solve the equation, factor x^{2}+3x-40 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.

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

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.

$$a=-5 b=8$$

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

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

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

$$x=5 x=-8$$

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

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

Simplify. The value x=5 satisfies the equation.

$$5=5$$

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

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

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

$$-8=8$$

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

$$x=5$$