Solve 4 * (2x - 1) ^ 2 = (x + 1) ^ 2 | 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

$$4\cdot(2x-1)^{2}=(x+1)^{2}$$

Solve for x

$x=\frac{1}{5}=0.2$
$x=1$

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring By Grouping

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

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

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

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

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

$$16x^{2}-16x+4=x^{2}+2x+1$$

Subtract $x^{2}$ from both sides.

$$16x^{2}-16x+4-x^{2}=2x+1$$

Combine $16x^{2}$ and $-x^{2}$ to get $15x^{2}$.

$$15x^{2}-16x+4=2x+1$$

Subtract $2x$ from both sides.

$$15x^{2}-16x+4-2x=1$$

Combine $-16x$ and $-2x$ to get $-18x$.

$$15x^{2}-18x+4=1$$

Subtract $1$ from both sides.

$$15x^{2}-18x+4-1=0$$

Subtract $1$ from $4$ to get $3$.

$$15x^{2}-18x+3=0$$

Divide both sides by $3$.

$$5x^{2}-6x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $5x^{2}+ax+bx+1$. To find $a$ and $b$, set up a system to be solved.

$$a+b=-6$$ $$ab=5\times 1=5$$

Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. The only such pair is the system solution.

$$a=-5$$ $$b=-1$$

Rewrite $5x^{2}-6x+1$ as $\left(5x^{2}-5x\right)+\left(-x+1\right)$.

$$\left(5x^{2}-5x\right)+\left(-x+1\right)$$

Factor out $5x$ in the first and $-1$ in the second group.

$$5x\left(x-1\right)-\left(x-1\right)$$

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

$$\left(x-1\right)\left(5x-1\right)$$

To find equation solutions, solve $x-1=0$ and $5x-1=0$.

$$x=1$$ $$x=\frac{1}{5}$$

Steps Using the Quadratic Formula

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

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

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

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

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

$$16x^{2}-16x+4=x^{2}+2x+1$$

Subtract $x^{2}$ from both sides.

$$16x^{2}-16x+4-x^{2}=2x+1$$

Combine $16x^{2}$ and $-x^{2}$ to get $15x^{2}$.

$$15x^{2}-16x+4=2x+1$$

Subtract $2x$ from both sides.

$$15x^{2}-16x+4-2x=1$$

Combine $-16x$ and $-2x$ to get $-18x$.

$$15x^{2}-18x+4=1$$

Subtract $1$ from both sides.

$$15x^{2}-18x+4-1=0$$

Subtract $1$ from $4$ to get $3$.

$$15x^{2}-18x+3=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $15$ for $a$, $-18$ for $b$, and $3$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 15\times 3}}{2\times 15}$$

Square $-18$.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\times 15\times 3}}{2\times 15}$$

Multiply $-4$ times $15$.

$$x=\frac{-\left(-18\right)±\sqrt{324-60\times 3}}{2\times 15}$$

Multiply $-60$ times $3$.

$$x=\frac{-\left(-18\right)±\sqrt{324-180}}{2\times 15}$$

Add $324$ to $-180$.

$$x=\frac{-\left(-18\right)±\sqrt{144}}{2\times 15}$$

Take the square root of $144$.

$$x=\frac{-\left(-18\right)±12}{2\times 15}$$

The opposite of $-18$ is $18$.

$$x=\frac{18±12}{2\times 15}$$

Multiply $2$ times $15$.

$$x=\frac{18±12}{30}$$

Now solve the equation $x=\frac{18±12}{30}$ when $±$ is plus. Add $18$ to $12$.

$$x=\frac{30}{30}$$

Divide $30$ by $30$.

$$x=1$$

Now solve the equation $x=\frac{18±12}{30}$ when $±$ is minus. Subtract $12$ from $18$.

$$x=\frac{6}{30}$$

Reduce the fraction $\frac{6}{30}$ to lowest terms by extracting and canceling out $6$.

$$x=\frac{1}{5}$$

The equation is now solved.

$$x=1$$ $$x=\frac{1}{5}$$

Steps for Completing the Square

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

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

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

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

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

$$16x^{2}-16x+4=x^{2}+2x+1$$

Subtract $x^{2}$ from both sides.

$$16x^{2}-16x+4-x^{2}=2x+1$$

Combine $16x^{2}$ and $-x^{2}$ to get $15x^{2}$.

$$15x^{2}-16x+4=2x+1$$

Subtract $2x$ from both sides.

$$15x^{2}-16x+4-2x=1$$

Combine $-16x$ and $-2x$ to get $-18x$.

$$15x^{2}-18x+4=1$$

Subtract $4$ from both sides.

$$15x^{2}-18x=1-4$$

Subtract $4$ from $1$ to get $-3$.

$$15x^{2}-18x=-3$$

Divide both sides by $15$.

$$\frac{15x^{2}-18x}{15}=-\frac{3}{15}$$

Dividing by $15$ undoes the multiplication by $15$.

$$x^{2}+\left(-\frac{18}{15}\right)x=-\frac{3}{15}$$

Reduce the fraction $\frac{-18}{15}$ to lowest terms by extracting and canceling out $3$.

$$x^{2}-\frac{6}{5}x=-\frac{3}{15}$$

Reduce the fraction $\frac{-3}{15}$ to lowest terms by extracting and canceling out $3$.

$$x^{2}-\frac{6}{5}x=-\frac{1}{5}$$

Divide $-\frac{6}{5}$, the coefficient of the $x$ term, by $2$ to get $-\frac{3}{5}$. Then add the square of $-\frac{3}{5}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{6}{5}x+\left(-\frac{3}{5}\right)^{2}=-\frac{1}{5}+\left(-\frac{3}{5}\right)^{2}$$

Square $-\frac{3}{5}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{6}{5}x+\frac{9}{25}=-\frac{1}{5}+\frac{9}{25}$$

Add $-\frac{1}{5}$ to $\frac{9}{25}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{6}{5}x+\frac{9}{25}=\frac{4}{25}$$

Factor $x^{2}-\frac{6}{5}x+\frac{9}{25}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{3}{5}\right)^{2}=\frac{4}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{5}\right)^{2}}=\sqrt{\frac{4}{25}}$$

Simplify.

$$x-\frac{3}{5}=\frac{2}{5}$$ $$x-\frac{3}{5}=-\frac{2}{5}$$

Add $\frac{3}{5}$ to both sides of the equation.

$$x=1$$ $$x=\frac{1}{5}$$