Solve (x)^(2)=(2x)/(121) | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$${ x }^{ 2 } = \frac{ 2x }{ 121 }$$

Solve for x

$x=\frac{2}{121}\approx 0.016528926$
$x=0$

Steps Using Factoring

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring

Multiply both sides of the equation by $121$.

$$121x^{2}=2x$$

Subtract $2x$ from both sides.

$$121x^{2}-2x=0$$

Factor out $x$.

$$x\left(121x-2\right)=0$$

To find equation solutions, solve $x=0$ and $121x-2=0$.

$$x=0$$ $$x=\frac{2}{121}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by $121$.

$$121x^{2}=2x$$

Subtract $2x$ from both sides.

$$121x^{2}-2x=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $121$ for $a$, $-2$ for $b$, and $0$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}}}{2\times 121}$$

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

$$x=\frac{-\left(-2\right)±2}{2\times 121}$$

The opposite of $-2$ is $2$.

$$x=\frac{2±2}{2\times 121}$$

Multiply $2$ times $121$.

$$x=\frac{2±2}{242}$$

Now solve the equation $x=\frac{2±2}{242}$ when $±$ is plus. Add $2$ to $2$.

$$x=\frac{4}{242}$$

Reduce the fraction $\frac{4}{242}$ to lowest terms by extracting and canceling out $2$.

$$x=\frac{2}{121}$$

Now solve the equation $x=\frac{2±2}{242}$ when $±$ is minus. Subtract $2$ from $2$.

$$x=\frac{0}{242}$$

Divide $0$ by $242$.

$$x=0$$

The equation is now solved.

$$x=\frac{2}{121}$$ $$x=0$$

Steps for Completing the Square

Multiply both sides of the equation by $121$.

$$121x^{2}=2x$$

Subtract $2x$ from both sides.

$$121x^{2}-2x=0$$

Divide both sides by $121$.

$$\frac{121x^{2}-2x}{121}=\frac{0}{121}$$

Dividing by $121$ undoes the multiplication by $121$.

$$x^{2}-\frac{2}{121}x=\frac{0}{121}$$

Divide $0$ by $121$.

$$x^{2}-\frac{2}{121}x=0$$

Divide $-\frac{2}{121}$, the coefficient of the $x$ term, by $2$ to get $-\frac{1}{121}$. Then add the square of $-\frac{1}{121}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{2}{121}x+\left(-\frac{1}{121}\right)^{2}=\left(-\frac{1}{121}\right)^{2}$$

Square $-\frac{1}{121}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{121}x+\frac{1}{14641}=\frac{1}{14641}$$

Factor $x^{2}-\frac{2}{121}x+\frac{1}{14641}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{1}{121}\right)^{2}=\frac{1}{14641}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{121}\right)^{2}}=\sqrt{\frac{1}{14641}}$$

Simplify.

$$x-\frac{1}{121}=\frac{1}{121}$$ $$x-\frac{1}{121}=-\frac{1}{121}$$

Add $\frac{1}{121}$ to both sides of the equation.

$$x=\frac{2}{121}$$ $$x=0$$