Solve 10xdiv1/4x=40x

$$10 x \div \frac { 1 } { 4 } x = 40 x$$

Solve for x

$$x=1$$

$$x=0$$

Steps Using Factoring

Divide 10x by \frac{1}{4} to get 40x.

$$40xx=40x$$

Multiply x and x to get x^{2}.

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

Subtract 40x from both sides.

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

Factor out x.

$$x\left(40x-40\right)=0$$

To find equation solutions, solve x=0 and 40x-40=0.

$$x=0 x=1$$

Steps Using the Quadratic Formula

Divide 10x by \frac{1}{4} to get 40x.

$$40xx=40x$$

Multiply x and x to get x^{2}.

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

Subtract 40x from both sides.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 40 for a, -40 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}}}{2\times 40}$$

Take the square root of \left(-40\right)^{2}.

$$x=\frac{-\left(-40\right)±40}{2\times 40}$$

The opposite of -40 is 40.

$$x=\frac{40±40}{2\times 40}$$

Multiply 2 times 40.

$$x=\frac{40±40}{80}$$

Now solve the equation x=\frac{40±40}{80} when ± is plus. Add 40 to 40.

$$x=\frac{80}{80}$$

Divide 80 by 80.

$$x=1$$

Now solve the equation x=\frac{40±40}{80} when ± is minus. Subtract 40 from 40.

$$x=\frac{0}{80}$$

Divide 0 by 80.

$$x=0$$

The equation is now solved.

$$x=1 x=0$$

Steps for Completing the Square

Divide 10x by \frac{1}{4} to get 40x.

$$40xx=40x$$

Multiply x and x to get x^{2}.

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

Subtract 40x from both sides.

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

Divide both sides by 40.

$$\frac{40x^{2}-40x}{40}=\frac{0}{40}$$

Dividing by 40 undoes the multiplication by 40.

$$x^{2}+\left(-\frac{40}{40}\right)x=\frac{0}{40}$$

Divide -40 by 40.

$$x^{2}-x=\frac{0}{40}$$

Divide 0 by 40.

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

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Factor x^{2}-x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

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

Add \frac{1}{2} to both sides of the equation.

$$x=1 x=0$$