Solve sqrt{2-7x}+2x=0

$$\sqrt{ 2-7x } +2x = 0$$

Solve for x

$$x=-2$$

Solution Steps

Subtract 2x from both sides of the equation.

$$\sqrt{2-7x}=-2x$$

Square both sides of the equation.

$$\left(\sqrt{2-7x}\right)^{2}=\left(-2x\right)^{2}$$

Calculate \sqrt{2-7x} to the power of 2 and get 2-7x.

$$2-7x=\left(-2x\right)^{2}$$

Expand \left(-2x\right)^{2}.

$$2-7x=\left(-2\right)^{2}x^{2}$$

Calculate -2 to the power of 2 and get 4.

$$2-7x=4x^{2}$$

Subtract 4x^{2} from both sides.

$$2-7x-4x^{2}=0$$

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

$$-4x^{2}-7x+2=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -4x^{2}+ax+bx+2. To find a and b, set up a system to be solved.

$$a+b=-7 ab=-4\times 2=-8$$

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 -8.

$$1,-8 2,-4$$

Calculate the sum for each pair.

$$1-8=-7 2-4=-2$$

The solution is the pair that gives sum -7.

$$a=1 b=-8$$

Rewrite -4x^{2}-7x+2 as \left(-4x^{2}+x\right)+\left(-8x+2\right).

$$\left(-4x^{2}+x\right)+\left(-8x+2\right)$$

Factor out -x in the first and -2 in the second group.

$$-x\left(4x-1\right)-2\left(4x-1\right)$$

Factor out common term 4x-1 by using distributive property.

$$\left(4x-1\right)\left(-x-2\right)$$

To find equation solutions, solve 4x-1=0 and -x-2=0.

$$x=\frac{1}{4} x=-2$$

Substitute \frac{1}{4} for x in the equation \sqrt{2-7x}+2x=0.

$$\sqrt{2-7\times \frac{1}{4}}+2\times \frac{1}{4}=0$$

Simplify. The value x=\frac{1}{4} does not satisfy the equation.

$$1=0$$

Substitute -2 for x in the equation \sqrt{2-7x}+2x=0.

$$\sqrt{2-7\left(-2\right)}+2\left(-2\right)=0$$

Simplify. The value x=-2 satisfies the equation.

$$0=0$$

Equation \sqrt{2-7x}=-2x has a unique solution.

$$x=-2$$