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

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

Solve for x

$x=\frac{3}{5}=0.6$
$x=1$

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

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

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

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

Subtract $4$ from both sides.

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

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

$$-3-4x+4x^{2}=-12x+9x^{2}$$

Add $12x$ to both sides.

$$-3-4x+4x^{2}+12x=9x^{2}$$

Combine $-4x$ and $12x$ to get $8x$.

$$-3+8x+4x^{2}=9x^{2}$$

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

$$-3+8x+4x^{2}-9x^{2}=0$$

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

$$-3+8x-5x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-5x^{2}+8x-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $-5x^{2}+ax+bx-3$. To find $a$ and $b$, set up a system to be solved.

$$a+b=8$$ $$ab=-5\left(-3\right)=15$$

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 $15$.

$$1,15$$ $$3,5$$

Calculate the sum for each pair.

$$1+15=16$$ $$3+5=8$$

The solution is the pair that gives sum $8$.

$$a=5$$ $$b=3$$

Rewrite $-5x^{2}+8x-3$ as $\left(-5x^{2}+5x\right)+\left(3x-3\right)$.

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

Factor out $5x$ in the first and $-3$ in the second group.

$$5x\left(-x+1\right)-3\left(-x+1\right)$$

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

$$\left(-x+1\right)\left(5x-3\right)$$

To find equation solutions, solve $-x+1=0$ and $5x-3=0$.

$$x=1$$ $$x=\frac{3}{5}$$

Steps Using the Quadratic Formula

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

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

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

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

Subtract $4$ from both sides.

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

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

$$-3-4x+4x^{2}=-12x+9x^{2}$$

Add $12x$ to both sides.

$$-3-4x+4x^{2}+12x=9x^{2}$$

Combine $-4x$ and $12x$ to get $8x$.

$$-3+8x+4x^{2}=9x^{2}$$

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

$$-3+8x+4x^{2}-9x^{2}=0$$

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

$$-3+8x-5x^{2}=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.

$$-5x^{2}+8x-3=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-5$ for $a$, $8$ for $b$, and $-3$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-5\right)\left(-3\right)}}{2\left(-5\right)}$$

Square $8$.

$$x=\frac{-8±\sqrt{64-4\left(-5\right)\left(-3\right)}}{2\left(-5\right)}$$

Multiply $-4$ times $-5$.

$$x=\frac{-8±\sqrt{64+20\left(-3\right)}}{2\left(-5\right)}$$

Multiply $20$ times $-3$.

$$x=\frac{-8±\sqrt{64-60}}{2\left(-5\right)}$$

Add $64$ to $-60$.

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

Take the square root of $4$.

$$x=\frac{-8±2}{2\left(-5\right)}$$

Multiply $2$ times $-5$.

$$x=\frac{-8±2}{-10}$$

Now solve the equation $x=\frac{-8±2}{-10}$ when $±$ is plus. Add $-8$ to $2$.

$$x=-\frac{6}{-10}$$

Reduce the fraction $\frac{-6}{-10}$ to lowest terms by extracting and canceling out $2$.

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

Now solve the equation $x=\frac{-8±2}{-10}$ when $±$ is minus. Subtract $2$ from $-8$.

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

Divide $-10$ by $-10$.

$$x=1$$

The equation is now solved.

$$x=\frac{3}{5}$$ $$x=1$$

Steps for Completing the Square

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

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

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

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

Add $12x$ to both sides.

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

Combine $-4x$ and $12x$ to get $8x$.

$$1+8x+4x^{2}=4+9x^{2}$$

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

$$1+8x+4x^{2}-9x^{2}=4$$

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

$$1+8x-5x^{2}=4$$

Subtract $1$ from both sides.

$$8x-5x^{2}=4-1$$

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

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.

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

Divide both sides by $-5$.

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

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

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

Divide $8$ by $-5$.

$$x^{2}-\frac{8}{5}x=\frac{3}{-5}$$

Divide $3$ by $-5$.

$$x^{2}-\frac{8}{5}x=-\frac{3}{5}$$

Divide $-\frac{8}{5}$, the coefficient of the $x$ term, by $2$ to get $-\frac{4}{5}$. Then add the square of $-\frac{4}{5}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{8}{5}x+\left(-\frac{4}{5}\right)^{2}=-\frac{3}{5}+\left(-\frac{4}{5}\right)^{2}$$

Square $-\frac{4}{5}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{8}{5}x+\frac{16}{25}=-\frac{3}{5}+\frac{16}{25}$$

Add $-\frac{3}{5}$ to $\frac{16}{25}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{8}{5}x+\frac{16}{25}=\frac{1}{25}$$

Factor $x^{2}-\frac{8}{5}x+\frac{16}{25}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{4}{5}\right)^{2}=\frac{1}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{4}{5}\right)^{2}}=\sqrt{\frac{1}{25}}$$

Simplify.

$$x-\frac{4}{5}=\frac{1}{5}$$ $$x-\frac{4}{5}=-\frac{1}{5}$$

Add $\frac{4}{5}$ to both sides of the equation.

$$x=1$$ $$x=\frac{3}{5}$$