Solve x * (2x - 1) ^ 2 - x(2x - 1)(2x + 1) = 0 | 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

$$x(2x-1)^{2}-x(2x-1)(2x+1)=0$$

Solve for x

$x=\frac{1}{2}=0.5$
$x=0$

Steps Using Factoring

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(2x-1\right)^{2}$.

$$x\left(4x^{2}-4x+1\right)-x\left(2x-1\right)\left(2x+1\right)=0$$

Use the distributive property to multiply $x$ by $4x^{2}-4x+1$.

$$4x^{3}-4x^{2}+x-x\left(2x-1\right)\left(2x+1\right)=0$$

Use the distributive property to multiply $x$ by $2x-1$.

$$4x^{3}-4x^{2}+x-\left(2x^{2}-x\right)\left(2x+1\right)=0$$

Use the distributive property to multiply $2x^{2}-x$ by $2x+1$ and combine like terms.

$$4x^{3}-4x^{2}+x-\left(4x^{3}-x\right)=0$$

To find the opposite of $4x^{3}-x$, find the opposite of each term.

$$4x^{3}-4x^{2}+x-4x^{3}+x=0$$

Combine $4x^{3}$ and $-4x^{3}$ to get $0$.

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

Combine $x$ and $x$ to get $2x$.

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

Factor out $x$.

$$x\left(-4x+2\right)=0$$

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

$$x=0$$ $$x=\frac{1}{2}$$

Steps Using the Quadratic Formula

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(2x-1\right)^{2}$.

$$x\left(4x^{2}-4x+1\right)-x\left(2x-1\right)\left(2x+1\right)=0$$

Use the distributive property to multiply $x$ by $4x^{2}-4x+1$.

$$4x^{3}-4x^{2}+x-x\left(2x-1\right)\left(2x+1\right)=0$$

Use the distributive property to multiply $x$ by $2x-1$.

$$4x^{3}-4x^{2}+x-\left(2x^{2}-x\right)\left(2x+1\right)=0$$

Use the distributive property to multiply $2x^{2}-x$ by $2x+1$ and combine like terms.

$$4x^{3}-4x^{2}+x-\left(4x^{3}-x\right)=0$$

To find the opposite of $4x^{3}-x$, find the opposite of each term.

$$4x^{3}-4x^{2}+x-4x^{3}+x=0$$

Combine $4x^{3}$ and $-4x^{3}$ to get $0$.

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

Combine $x$ and $x$ to get $2x$.

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

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

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

Take the square root of $2^{2}$.

$$x=\frac{-2±2}{2\left(-4\right)}$$

Multiply $2$ times $-4$.

$$x=\frac{-2±2}{-8}$$

Now solve the equation $x=\frac{-2±2}{-8}$ when $±$ is plus. Add $-2$ to $2$.

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

Divide $0$ by $-8$.

$$x=0$$

Now solve the equation $x=\frac{-2±2}{-8}$ when $±$ is minus. Subtract $2$ from $-2$.

$$x=-\frac{4}{-8}$$

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

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

The equation is now solved.

$$x=0$$ $$x=\frac{1}{2}$$

Steps for Completing the Square

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(2x-1\right)^{2}$.

$$x\left(4x^{2}-4x+1\right)-x\left(2x-1\right)\left(2x+1\right)=0$$

Use the distributive property to multiply $x$ by $4x^{2}-4x+1$.

$$4x^{3}-4x^{2}+x-x\left(2x-1\right)\left(2x+1\right)=0$$

Use the distributive property to multiply $x$ by $2x-1$.

$$4x^{3}-4x^{2}+x-\left(2x^{2}-x\right)\left(2x+1\right)=0$$

Use the distributive property to multiply $2x^{2}-x$ by $2x+1$ and combine like terms.

$$4x^{3}-4x^{2}+x-\left(4x^{3}-x\right)=0$$

To find the opposite of $4x^{3}-x$, find the opposite of each term.

$$4x^{3}-4x^{2}+x-4x^{3}+x=0$$

Combine $4x^{3}$ and $-4x^{3}$ to get $0$.

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

Combine $x$ and $x$ to get $2x$.

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

Divide both sides by $-4$.

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

Dividing by $-4$ undoes the multiplication by $-4$.

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

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

$$x^{2}-\frac{1}{2}x=\frac{0}{-4}$$

Divide $0$ by $-4$.

$$x^{2}-\frac{1}{2}x=0$$

Divide $-\frac{1}{2}$, the coefficient of the $x$ term, by $2$ to get $-\frac{1}{4}$. Then add the square of $-\frac{1}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square $-\frac{1}{4}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{1}{16}$$

Factor $x^{2}-\frac{1}{2}x+\frac{1}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

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

Add $\frac{1}{4}$ to both sides of the equation.

$$x=\frac{1}{2}$$ $$x=0$$