Solve (3x-2)(x-1)=4(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

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

Solve for x

$x=2$
$x=1$

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring By Grouping

Use the distributive property to multiply $3x-2$ by $x-1$ and combine like terms.

$$3x^{2}-5x+2=4\left(x-1\right)^{2}$$

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

$$3x^{2}-5x+2=4\left(x^{2}-2x+1\right)$$

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

$$3x^{2}-5x+2=4x^{2}-8x+4$$

Subtract $4x^{2}$ from both sides.

$$3x^{2}-5x+2-4x^{2}=-8x+4$$

Combine $3x^{2}$ and $-4x^{2}$ to get $-x^{2}$.

$$-x^{2}-5x+2=-8x+4$$

Add $8x$ to both sides.

$$-x^{2}-5x+2+8x=4$$

Combine $-5x$ and $8x$ to get $3x$.

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

Subtract $4$ from both sides.

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

Subtract $4$ from $2$ to get $-2$.

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

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

$$a+b=3$$ $$ab=-\left(-2\right)=2$$

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

$$a=2$$ $$b=1$$

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

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

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

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

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

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

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

$$x=2$$ $$x=1$$

Steps Using the Quadratic Formula

Use the distributive property to multiply $3x-2$ by $x-1$ and combine like terms.

$$3x^{2}-5x+2=4\left(x-1\right)^{2}$$

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

$$3x^{2}-5x+2=4\left(x^{2}-2x+1\right)$$

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

$$3x^{2}-5x+2=4x^{2}-8x+4$$

Subtract $4x^{2}$ from both sides.

$$3x^{2}-5x+2-4x^{2}=-8x+4$$

Combine $3x^{2}$ and $-4x^{2}$ to get $-x^{2}$.

$$-x^{2}-5x+2=-8x+4$$

Add $8x$ to both sides.

$$-x^{2}-5x+2+8x=4$$

Combine $-5x$ and $8x$ to get $3x$.

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

Subtract $4$ from both sides.

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

Subtract $4$ from $2$ to get $-2$.

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

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-1$ for $a$, $3$ for $b$, and $-2$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-1\right)\left(-2\right)}}{2\left(-1\right)}$$

Square $3$.

$$x=\frac{-3±\sqrt{9-4\left(-1\right)\left(-2\right)}}{2\left(-1\right)}$$

Multiply $-4$ times $-1$.

$$x=\frac{-3±\sqrt{9+4\left(-2\right)}}{2\left(-1\right)}$$

Multiply $4$ times $-2$.

$$x=\frac{-3±\sqrt{9-8}}{2\left(-1\right)}$$

Add $9$ to $-8$.

$$x=\frac{-3±\sqrt{1}}{2\left(-1\right)}$$

Take the square root of $1$.

$$x=\frac{-3±1}{2\left(-1\right)}$$

Multiply $2$ times $-1$.

$$x=\frac{-3±1}{-2}$$

Now solve the equation $x=\frac{-3±1}{-2}$ when $±$ is plus. Add $-3$ to $1$.

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

Divide $-2$ by $-2$.

$$x=1$$

Now solve the equation $x=\frac{-3±1}{-2}$ when $±$ is minus. Subtract $1$ from $-3$.

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

Divide $-4$ by $-2$.

$$x=2$$

The equation is now solved.

$$x=1$$ $$x=2$$

Steps for Completing the Square

Use the distributive property to multiply $3x-2$ by $x-1$ and combine like terms.

$$3x^{2}-5x+2=4\left(x-1\right)^{2}$$

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

$$3x^{2}-5x+2=4\left(x^{2}-2x+1\right)$$

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

$$3x^{2}-5x+2=4x^{2}-8x+4$$

Subtract $4x^{2}$ from both sides.

$$3x^{2}-5x+2-4x^{2}=-8x+4$$

Combine $3x^{2}$ and $-4x^{2}$ to get $-x^{2}$.

$$-x^{2}-5x+2=-8x+4$$

Add $8x$ to both sides.

$$-x^{2}-5x+2+8x=4$$

Combine $-5x$ and $8x$ to get $3x$.

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

Subtract $2$ from both sides.

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

Subtract $2$ from $4$ to get $2$.

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

Divide both sides by $-1$.

$$\frac{-x^{2}+3x}{-1}=\frac{2}{-1}$$

Dividing by $-1$ undoes the multiplication by $-1$.

$$x^{2}+\frac{3}{-1}x=\frac{2}{-1}$$

Divide $3$ by $-1$.

$$x^{2}-3x=\frac{2}{-1}$$

Divide $2$ by $-1$.

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

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

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

Square $-\frac{3}{2}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=-2+\frac{9}{4}$$

Add $-2$ to $\frac{9}{4}$.

$$x^{2}-3x+\frac{9}{4}=\frac{1}{4}$$

Factor $x^{2}-3x+\frac{9}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add $\frac{3}{2}$ to both sides of the equation.

$$x=2$$ $$x=1$$