Solve (3x - 1) ^ 2 - 25x ^ 2 = 0 | 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

$$(3x-1)^{2}-25x^{2}=0$$

Solve for x

$x=\frac{1}{8}=0.125$
$x=-\frac{1}{2}=-0.5$

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

$$9x^{2}-6x+1-25x^{2}=0$$

Combine $9x^{2}$ and $-25x^{2}$ to get $-16x^{2}$.

$$-16x^{2}-6x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $-16x^{2}+ax+bx+1$. To find $a$ and $b$, set up a system to be solved.

$$a+b=-6$$ $$ab=-16=-16$$

Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $-16$.

$$1,-16$$ $$2,-8$$ $$4,-4$$

Calculate the sum for each pair.

$$1-16=-15$$ $$2-8=-6$$ $$4-4=0$$

The solution is the pair that gives sum $-6$.

$$a=2$$ $$b=-8$$

Rewrite $-16x^{2}-6x+1$ as $\left(-16x^{2}+2x\right)+\left(-8x+1\right)$.

$$\left(-16x^{2}+2x\right)+\left(-8x+1\right)$$

Factor out $2x$ in $-16x^{2}+2x$.

$$2x\left(-8x+1\right)-8x+1$$

Factor out common term $-8x+1$ by using distributive property.

$$\left(-8x+1\right)\left(2x+1\right)$$

To find equation solutions, solve $-8x+1=0$ and $2x+1=0$.

$$x=\frac{1}{8}$$ $$x=-\frac{1}{2}$$

Steps Using the Quadratic Formula

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

$$9x^{2}-6x+1-25x^{2}=0$$

Combine $9x^{2}$ and $-25x^{2}$ to get $-16x^{2}$.

$$-16x^{2}-6x+1=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-16$ for $a$, $-6$ for $b$, and $1$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-16\right)}}{2\left(-16\right)}$$

Square $-6$.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\left(-16\right)}}{2\left(-16\right)}$$

Multiply $-4$ times $-16$.

$$x=\frac{-\left(-6\right)±\sqrt{36+64}}{2\left(-16\right)}$$

Add $36$ to $64$.

$$x=\frac{-\left(-6\right)±\sqrt{100}}{2\left(-16\right)}$$

Take the square root of $100$.

$$x=\frac{-\left(-6\right)±10}{2\left(-16\right)}$$

The opposite of $-6$ is $6$.

$$x=\frac{6±10}{2\left(-16\right)}$$

Multiply $2$ times $-16$.

$$x=\frac{6±10}{-32}$$

Now solve the equation $x=\frac{6±10}{-32}$ when $±$ is plus. Add $6$ to $10$.

$$x=\frac{16}{-32}$$

Reduce the fraction $\frac{16}{-32}$ to lowest terms by extracting and canceling out $16$.

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

Now solve the equation $x=\frac{6±10}{-32}$ when $±$ is minus. Subtract $10$ from $6$.

$$x=-\frac{4}{-32}$$

Reduce the fraction $\frac{-4}{-32}$ to lowest terms by extracting and canceling out $4$.

$$x=\frac{1}{8}$$

The equation is now solved.

$$x=-\frac{1}{2}$$ $$x=\frac{1}{8}$$

Steps for Completing the Square

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

$$9x^{2}-6x+1-25x^{2}=0$$

Combine $9x^{2}$ and $-25x^{2}$ to get $-16x^{2}$.

$$-16x^{2}-6x+1=0$$

Subtract $1$ from both sides. Anything subtracted from zero gives its negation.

$$-16x^{2}-6x=-1$$

Divide both sides by $-16$.

$$\frac{-16x^{2}-6x}{-16}=-\frac{1}{-16}$$

Dividing by $-16$ undoes the multiplication by $-16$.

$$x^{2}+\left(-\frac{6}{-16}\right)x=-\frac{1}{-16}$$

Reduce the fraction $\frac{-6}{-16}$ to lowest terms by extracting and canceling out $2$.

$$x^{2}+\frac{3}{8}x=-\frac{1}{-16}$$

Divide $-1$ by $-16$.

$$x^{2}+\frac{3}{8}x=\frac{1}{16}$$

Divide $\frac{3}{8}$, the coefficient of the $x$ term, by $2$ to get $\frac{3}{16}$. Then add the square of $\frac{3}{16}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{8}x+\left(\frac{3}{16}\right)^{2}=\frac{1}{16}+\left(\frac{3}{16}\right)^{2}$$

Square $\frac{3}{16}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{8}x+\frac{9}{256}=\frac{1}{16}+\frac{9}{256}$$

Add $\frac{1}{16}$ to $\frac{9}{256}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{3}{8}x+\frac{9}{256}=\frac{25}{256}$$

Factor $x^{2}+\frac{3}{8}x+\frac{9}{256}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{3}{16}\right)^{2}=\frac{25}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{16}\right)^{2}}=\sqrt{\frac{25}{256}}$$

Simplify.

$$x+\frac{3}{16}=\frac{5}{16}$$ $$x+\frac{3}{16}=-\frac{5}{16}$$

Subtract $\frac{3}{16}$ from both sides of the equation.

$$x=\frac{1}{8}$$ $$x=-\frac{1}{2}$$