Solve (x + 1) ^ 2 = (3x - 2) ^ 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

$$(x+1)^{2}=(3x-2)^{2}$$

Solve for x

$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$
$x=\frac{1}{4}=0.25$

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

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

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(3x-2\right)^{2}$.

$$x^{2}+2x+1=9x^{2}-12x+4$$

Subtract $9x^{2}$ from both sides.

$$x^{2}+2x+1-9x^{2}=-12x+4$$

Combine $x^{2}$ and $-9x^{2}$ to get $-8x^{2}$.

$$-8x^{2}+2x+1=-12x+4$$

Add $12x$ to both sides.

$$-8x^{2}+2x+1+12x=4$$

Combine $2x$ and $12x$ to get $14x$.

$$-8x^{2}+14x+1=4$$

Subtract $4$ from both sides.

$$-8x^{2}+14x+1-4=0$$

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

$$-8x^{2}+14x-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $-8x^{2}+ax+bx-3$. To find $a$ and $b$, set up a system to be solved.

$$a+b=14$$ $$ab=-8\left(-3\right)=24$$

Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. List all such integer pairs that give product $24$.

$$1,24$$ $$2,12$$ $$3,8$$ $$4,6$$

Calculate the sum for each pair.

$$1+24=25$$ $$2+12=14$$ $$3+8=11$$ $$4+6=10$$

The solution is the pair that gives sum $14$.

$$a=12$$ $$b=2$$

Rewrite $-8x^{2}+14x-3$ as $\left(-8x^{2}+12x\right)+\left(2x-3\right)$.

$$\left(-8x^{2}+12x\right)+\left(2x-3\right)$$

Factor out $-4x$ in $-8x^{2}+12x$.

$$-4x\left(2x-3\right)+2x-3$$

Factor out common term $2x-3$ by using distributive property.

$$\left(2x-3\right)\left(-4x+1\right)$$

To find equation solutions, solve $2x-3=0$ and $-4x+1=0$.

$$x=\frac{3}{2}$$ $$x=\frac{1}{4}$$

Steps Using the Quadratic Formula

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

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

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(3x-2\right)^{2}$.

$$x^{2}+2x+1=9x^{2}-12x+4$$

Subtract $9x^{2}$ from both sides.

$$x^{2}+2x+1-9x^{2}=-12x+4$$

Combine $x^{2}$ and $-9x^{2}$ to get $-8x^{2}$.

$$-8x^{2}+2x+1=-12x+4$$

Add $12x$ to both sides.

$$-8x^{2}+2x+1+12x=4$$

Combine $2x$ and $12x$ to get $14x$.

$$-8x^{2}+14x+1=4$$

Subtract $4$ from both sides.

$$-8x^{2}+14x+1-4=0$$

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

$$-8x^{2}+14x-3=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-8$ for $a$, $14$ for $b$, and $-3$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-14±\sqrt{14^{2}-4\left(-8\right)\left(-3\right)}}{2\left(-8\right)}$$

Square $14$.

$$x=\frac{-14±\sqrt{196-4\left(-8\right)\left(-3\right)}}{2\left(-8\right)}$$

Multiply $-4$ times $-8$.

$$x=\frac{-14±\sqrt{196+32\left(-3\right)}}{2\left(-8\right)}$$

Multiply $32$ times $-3$.

$$x=\frac{-14±\sqrt{196-96}}{2\left(-8\right)}$$

Add $196$ to $-96$.

$$x=\frac{-14±\sqrt{100}}{2\left(-8\right)}$$

Take the square root of $100$.

$$x=\frac{-14±10}{2\left(-8\right)}$$

Multiply $2$ times $-8$.

$$x=\frac{-14±10}{-16}$$

Now solve the equation $x=\frac{-14±10}{-16}$ when $±$ is plus. Add $-14$ to $10$.

$$x=-\frac{4}{-16}$$

Reduce the fraction $\frac{-4}{-16}$ to lowest terms by extracting and canceling out $4$.

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

Now solve the equation $x=\frac{-14±10}{-16}$ when $±$ is minus. Subtract $10$ from $-14$.

$$x=-\frac{24}{-16}$$

Reduce the fraction $\frac{-24}{-16}$ to lowest terms by extracting and canceling out $8$.

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

The equation is now solved.

$$x=\frac{1}{4}$$ $$x=\frac{3}{2}$$

Steps for Completing the Square

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

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

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(3x-2\right)^{2}$.

$$x^{2}+2x+1=9x^{2}-12x+4$$

Subtract $9x^{2}$ from both sides.

$$x^{2}+2x+1-9x^{2}=-12x+4$$

Combine $x^{2}$ and $-9x^{2}$ to get $-8x^{2}$.

$$-8x^{2}+2x+1=-12x+4$$

Add $12x$ to both sides.

$$-8x^{2}+2x+1+12x=4$$

Combine $2x$ and $12x$ to get $14x$.

$$-8x^{2}+14x+1=4$$

Subtract $1$ from both sides.

$$-8x^{2}+14x=4-1$$

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

$$-8x^{2}+14x=3$$

Divide both sides by $-8$.

$$\frac{-8x^{2}+14x}{-8}=\frac{3}{-8}$$

Dividing by $-8$ undoes the multiplication by $-8$.

$$x^{2}+\frac{14}{-8}x=\frac{3}{-8}$$

Reduce the fraction $\frac{14}{-8}$ to lowest terms by extracting and canceling out $2$.

$$x^{2}-\frac{7}{4}x=\frac{3}{-8}$$

Divide $3$ by $-8$.

$$x^{2}-\frac{7}{4}x=-\frac{3}{8}$$

Divide $-\frac{7}{4}$, the coefficient of the $x$ term, by $2$ to get $-\frac{7}{8}$. Then add the square of $-\frac{7}{8}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{4}x+\left(-\frac{7}{8}\right)^{2}=-\frac{3}{8}+\left(-\frac{7}{8}\right)^{2}$$

Square $-\frac{7}{8}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{4}x+\frac{49}{64}=-\frac{3}{8}+\frac{49}{64}$$

Add $-\frac{3}{8}$ to $\frac{49}{64}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{4}x+\frac{49}{64}=\frac{25}{64}$$

Factor $x^{2}-\frac{7}{4}x+\frac{49}{64}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{7}{8}\right)^{2}=\frac{25}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{8}\right)^{2}}=\sqrt{\frac{25}{64}}$$

Simplify.

$$x-\frac{7}{8}=\frac{5}{8}$$ $$x-\frac{7}{8}=-\frac{5}{8}$$

Add $\frac{7}{8}$ to both sides of the equation.

$$x=\frac{3}{2}$$ $$x=\frac{1}{4}$$