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

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

Solve for x

$x=4$
$x=-\frac{2}{5}=-0.4$

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 $\left(2x+3\right)^{2}$.

$$4x^{2}+12x+9=\left(3x-1\right)^{2}$$

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(3x-1\right)^{2}$.

$$4x^{2}+12x+9=9x^{2}-6x+1$$

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

$$4x^{2}+12x+9-9x^{2}=-6x+1$$

Combine $4x^{2}$ and $-9x^{2}$ to get $-5x^{2}$.

$$-5x^{2}+12x+9=-6x+1$$

Add $6x$ to both sides.

$$-5x^{2}+12x+9+6x=1$$

Combine $12x$ and $6x$ to get $18x$.

$$-5x^{2}+18x+9=1$$

Subtract $1$ from both sides.

$$-5x^{2}+18x+9-1=0$$

Subtract $1$ from $9$ to get $8$.

$$-5x^{2}+18x+8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $-5x^{2}+ax+bx+8$. To find $a$ and $b$, set up a system to be solved.

$$a+b=18$$ $$ab=-5\times 8=-40$$

Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product $-40$.

$$-1,40$$ $$-2,20$$ $$-4,10$$ $$-5,8$$

Calculate the sum for each pair.

$$-1+40=39$$ $$-2+20=18$$ $$-4+10=6$$ $$-5+8=3$$

The solution is the pair that gives sum $18$.

$$a=20$$ $$b=-2$$

Rewrite $-5x^{2}+18x+8$ as $\left(-5x^{2}+20x\right)+\left(-2x+8\right)$.

$$\left(-5x^{2}+20x\right)+\left(-2x+8\right)$$

Factor out $5x$ in the first and $2$ in the second group.

$$5x\left(-x+4\right)+2\left(-x+4\right)$$

Factor out common term $-x+4$ by using distributive property.

$$\left(-x+4\right)\left(5x+2\right)$$

To find equation solutions, solve $-x+4=0$ and $5x+2=0$.

$$x=4$$ $$x=-\frac{2}{5}$$

Steps Using the Quadratic Formula

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

$$4x^{2}+12x+9=\left(3x-1\right)^{2}$$

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(3x-1\right)^{2}$.

$$4x^{2}+12x+9=9x^{2}-6x+1$$

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

$$4x^{2}+12x+9-9x^{2}=-6x+1$$

Combine $4x^{2}$ and $-9x^{2}$ to get $-5x^{2}$.

$$-5x^{2}+12x+9=-6x+1$$

Add $6x$ to both sides.

$$-5x^{2}+12x+9+6x=1$$

Combine $12x$ and $6x$ to get $18x$.

$$-5x^{2}+18x+9=1$$

Subtract $1$ from both sides.

$$-5x^{2}+18x+9-1=0$$

Subtract $1$ from $9$ to get $8$.

$$-5x^{2}+18x+8=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-5$ for $a$, $18$ for $b$, and $8$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-18±\sqrt{18^{2}-4\left(-5\right)\times 8}}{2\left(-5\right)}$$

Square $18$.

$$x=\frac{-18±\sqrt{324-4\left(-5\right)\times 8}}{2\left(-5\right)}$$

Multiply $-4$ times $-5$.

$$x=\frac{-18±\sqrt{324+20\times 8}}{2\left(-5\right)}$$

Multiply $20$ times $8$.

$$x=\frac{-18±\sqrt{324+160}}{2\left(-5\right)}$$

Add $324$ to $160$.

$$x=\frac{-18±\sqrt{484}}{2\left(-5\right)}$$

Take the square root of $484$.

$$x=\frac{-18±22}{2\left(-5\right)}$$

Multiply $2$ times $-5$.

$$x=\frac{-18±22}{-10}$$

Now solve the equation $x=\frac{-18±22}{-10}$ when $±$ is plus. Add $-18$ to $22$.

$$x=\frac{4}{-10}$$

Reduce the fraction $\frac{4}{-10}$ to lowest terms by extracting and canceling out $2$.

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

Now solve the equation $x=\frac{-18±22}{-10}$ when $±$ is minus. Subtract $22$ from $-18$.

$$x=-\frac{40}{-10}$$

Divide $-40$ by $-10$.

$$x=4$$

The equation is now solved.

$$x=-\frac{2}{5}$$ $$x=4$$

Steps for Completing the Square

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

$$4x^{2}+12x+9=\left(3x-1\right)^{2}$$

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(3x-1\right)^{2}$.

$$4x^{2}+12x+9=9x^{2}-6x+1$$

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

$$4x^{2}+12x+9-9x^{2}=-6x+1$$

Combine $4x^{2}$ and $-9x^{2}$ to get $-5x^{2}$.

$$-5x^{2}+12x+9=-6x+1$$

Add $6x$ to both sides.

$$-5x^{2}+12x+9+6x=1$$

Combine $12x$ and $6x$ to get $18x$.

$$-5x^{2}+18x+9=1$$

Subtract $9$ from both sides.

$$-5x^{2}+18x=1-9$$

Subtract $9$ from $1$ to get $-8$.

$$-5x^{2}+18x=-8$$

Divide both sides by $-5$.

$$\frac{-5x^{2}+18x}{-5}=-\frac{8}{-5}$$

Dividing by $-5$ undoes the multiplication by $-5$.

$$x^{2}+\frac{18}{-5}x=-\frac{8}{-5}$$

Divide $18$ by $-5$.

$$x^{2}-\frac{18}{5}x=-\frac{8}{-5}$$

Divide $-8$ by $-5$.

$$x^{2}-\frac{18}{5}x=\frac{8}{5}$$

Divide $-\frac{18}{5}$, the coefficient of the $x$ term, by $2$ to get $-\frac{9}{5}$. Then add the square of $-\frac{9}{5}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{18}{5}x+\left(-\frac{9}{5}\right)^{2}=\frac{8}{5}+\left(-\frac{9}{5}\right)^{2}$$

Square $-\frac{9}{5}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{18}{5}x+\frac{81}{25}=\frac{8}{5}+\frac{81}{25}$$

Add $\frac{8}{5}$ to $\frac{81}{25}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{18}{5}x+\frac{81}{25}=\frac{121}{25}$$

Factor $x^{2}-\frac{18}{5}x+\frac{81}{25}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{9}{5}\right)^{2}=\frac{121}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{5}\right)^{2}}=\sqrt{\frac{121}{25}}$$

Simplify.

$$x-\frac{9}{5}=\frac{11}{5}$$ $$x-\frac{9}{5}=-\frac{11}{5}$$

Add $\frac{9}{5}$ to both sides of the equation.

$$x=4$$ $$x=-\frac{2}{5}$$