Solve 4x+2x/x^2=4

$$\frac { 4 x + 2 x } { x ^ { 2 } } = 4$$

Solve for x

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Steps Using Factoring

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x^{2}.

$$4x+2x=4x^{2}$$

Combine 4x and 2x to get 6x.

$$6x=4x^{2}$$

Subtract 4x^{2} from both sides.

$$6x-4x^{2}=0$$

Factor out x.

$$x\left(6-4x\right)=0$$

To find equation solutions, solve x=0 and 6-4x=0.

$$x=0 x=\frac{3}{2}$$

Variable x cannot be equal to 0.

$$x=\frac{3}{2}$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x^{2}.

$$4x+2x=4x^{2}$$

Combine 4x and 2x to get 6x.

$$6x=4x^{2}$$

Subtract 4x^{2} from both sides.

$$6x-4x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-4x^{2}+6x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 6 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}}}{2\left(-4\right)}$$

Take the square root of 6^{2}.

$$x=\frac{-6±6}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-6±6}{-8}$$

Now solve the equation x=\frac{-6±6}{-8} when ± is plus. Add -6 to 6.

$$x=\frac{0}{-8}$$

Divide 0 by -8.

$$x=0$$

Now solve the equation x=\frac{-6±6}{-8} when ± is minus. Subtract 6 from -6.

$$x=-\frac{12}{-8}$$

Reduce the fraction \frac{-12}{-8} to lowest terms by extracting and canceling out 4.

$$x=\frac{3}{2}$$

The equation is now solved.

$$x=0 x=\frac{3}{2}$$

Variable x cannot be equal to 0.

$$x=\frac{3}{2}$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x^{2}.

$$4x+2x=4x^{2}$$

Combine 4x and 2x to get 6x.

$$6x=4x^{2}$$

Subtract 4x^{2} from both sides.

$$6x-4x^{2}=0$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-4x^{2}+6x=0$$

Divide both sides by -4.

$$\frac{-4x^{2}+6x}{-4}=\frac{0}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{6}{-4}x=\frac{0}{-4}$$

Reduce the fraction \frac{6}{-4} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{3}{2}x=\frac{0}{-4}$$

Divide 0 by -4.

$$x^{2}-\frac{3}{2}x=0$$

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{9}{16}$$

Factor x^{2}-\frac{3}{2}x+\frac{9}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{3}{4} x-\frac{3}{4}=-\frac{3}{4}$$

Add \frac{3}{4} to both sides of the equation.

$$x=\frac{3}{2} x=0$$

Variable x cannot be equal to 0.

$$x=\frac{3}{2}$$