Solve (x ^ 2)/3 - x/12 = 1/24 | 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

$$\frac{x^{2}}{3}-\frac{x}{12}=\frac{1}{24}$$

Solve for x

$x=-\frac{1}{4}=-0.25$
$x=\frac{1}{2}=0.5$

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring By Grouping

Multiply both sides of the equation by $24$, the least common multiple of $3,12,24$.

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

Subtract $1$ from both sides.

$$8x^{2}-2x-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $8x^{2}+ax+bx-1$. To find $a$ and $b$, set up a system to be solved.

$$a+b=-2$$ $$ab=8\left(-1\right)=-8$$

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 $-8$.

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

Calculate the sum for each pair.

$$1-8=-7$$ $$2-4=-2$$

The solution is the pair that gives sum $-2$.

$$a=-4$$ $$b=2$$

Rewrite $8x^{2}-2x-1$ as $\left(8x^{2}-4x\right)+\left(2x-1\right)$.

$$\left(8x^{2}-4x\right)+\left(2x-1\right)$$

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

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

Factor out common term $2x-1$ by using distributive property.

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

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

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

Steps Using the Quadratic Formula

Multiply both sides of the equation by $24$, the least common multiple of $3,12,24$.

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

Subtract $1$ from both sides.

$$8x^{2}-2x-1=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $8$ for $a$, $-2$ for $b$, and $-1$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 8\left(-1\right)}}{2\times 8}$$

Square $-2$.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 8\left(-1\right)}}{2\times 8}$$

Multiply $-4$ times $8$.

$$x=\frac{-\left(-2\right)±\sqrt{4-32\left(-1\right)}}{2\times 8}$$

Multiply $-32$ times $-1$.

$$x=\frac{-\left(-2\right)±\sqrt{4+32}}{2\times 8}$$

Add $4$ to $32$.

$$x=\frac{-\left(-2\right)±\sqrt{36}}{2\times 8}$$

Take the square root of $36$.

$$x=\frac{-\left(-2\right)±6}{2\times 8}$$

The opposite of $-2$ is $2$.

$$x=\frac{2±6}{2\times 8}$$

Multiply $2$ times $8$.

$$x=\frac{2±6}{16}$$

Now solve the equation $x=\frac{2±6}{16}$ when $±$ is plus. Add $2$ to $6$.

$$x=\frac{8}{16}$$

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

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

Now solve the equation $x=\frac{2±6}{16}$ when $±$ is minus. Subtract $6$ from $2$.

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

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

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

The equation is now solved.

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

Steps for Completing the Square

Multiply both sides of the equation by $24$, the least common multiple of $3,12,24$.

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

Divide both sides by $8$.

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

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

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

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

$$x^{2}-\frac{1}{4}x=\frac{1}{8}$$

Divide $-\frac{1}{4}$, the coefficient of the $x$ term, by $2$ to get $-\frac{1}{8}$. Then add the square of $-\frac{1}{8}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{4}x+\left(-\frac{1}{8}\right)^{2}=\frac{1}{8}+\left(-\frac{1}{8}\right)^{2}$$

Square $-\frac{1}{8}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{4}x+\frac{1}{64}=\frac{1}{8}+\frac{1}{64}$$

Add $\frac{1}{8}$ to $\frac{1}{64}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{4}x+\frac{1}{64}=\frac{9}{64}$$

Factor $x^{2}-\frac{1}{4}x+\frac{1}{64}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{1}{8}\right)^{2}=\frac{9}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{8}\right)^{2}}=\sqrt{\frac{9}{64}}$$

Simplify.

$$x-\frac{1}{8}=\frac{3}{8}$$ $$x-\frac{1}{8}=-\frac{3}{8}$$

Add $\frac{1}{8}$ to both sides of the equation.

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