Solve 5/(4x) - (x + 2)/3 = x - 1 | Equatix - Math Solver

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Example

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3(r+2s)=2t-4

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Question

$$\frac{5}{4x}-\frac{x+2}{3}=x-1$$

Solve for x

$x = \frac{\sqrt{61} + 1}{8} \approx 1.101281209$
$x=\frac{1-\sqrt{61}}{8}\approx -0.851281209$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $12x$, the least common multiple of $4x,3$.

$$3\times 5-4x\left(x+2\right)=12xx+12x\left(-1\right)$$

Multiply $3$ and $5$ to get $15$.

$$15-4x\left(x+2\right)=12xx+12x\left(-1\right)$$

Use the distributive property to multiply $4x$ by $x+2$.

$$15-\left(4x^{2}+8x\right)=12xx+12x\left(-1\right)$$

To find the opposite of $4x^{2}+8x$, find the opposite of each term.

$$15-4x^{2}-8x=12xx+12x\left(-1\right)$$

Multiply $x$ and $x$ to get $x^{2}$.

$$15-4x^{2}-8x=12x^{2}+12x\left(-1\right)$$

Multiply $12$ and $-1$ to get $-12$.

$$15-4x^{2}-8x=12x^{2}-12x$$

Subtract $12x^{2}$ from both sides.

$$15-4x^{2}-8x-12x^{2}=-12x$$

Combine $-4x^{2}$ and $-12x^{2}$ to get $-16x^{2}$.

$$15-16x^{2}-8x=-12x$$

Add $12x$ to both sides.

$$15-16x^{2}-8x+12x=0$$

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

$$15-16x^{2}+4x=0$$

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.

$$-16x^{2}+4x+15=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-16$ for $a$, $4$ for $b$, and $15$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-4±\sqrt{4^{2}-4\left(-16\right)\times 15}}{2\left(-16\right)}$$

Square $4$.

$$x=\frac{-4±\sqrt{16-4\left(-16\right)\times 15}}{2\left(-16\right)}$$

Multiply $-4$ times $-16$.

$$x=\frac{-4±\sqrt{16+64\times 15}}{2\left(-16\right)}$$

Multiply $64$ times $15$.

$$x=\frac{-4±\sqrt{16+960}}{2\left(-16\right)}$$

Add $16$ to $960$.

$$x=\frac{-4±\sqrt{976}}{2\left(-16\right)}$$

Take the square root of $976$.

$$x=\frac{-4±4\sqrt{61}}{2\left(-16\right)}$$

Multiply $2$ times $-16$.

$$x=\frac{-4±4\sqrt{61}}{-32}$$

Now solve the equation $x=\frac{-4±4\sqrt{61}}{-32}$ when $±$ is plus. Add $-4$ to $4\sqrt{61}$.

$$x=\frac{4\sqrt{61}-4}{-32}$$

Divide $-4+4\sqrt{61}$ by $-32$.

$$x=\frac{1-\sqrt{61}}{8}$$

Now solve the equation $x=\frac{-4±4\sqrt{61}}{-32}$ when $±$ is minus. Subtract $4\sqrt{61}$ from $-4$.

$$x=\frac{-4\sqrt{61}-4}{-32}$$

Divide $-4-4\sqrt{61}$ by $-32$.

$$x=\frac{\sqrt{61}+1}{8}$$

The equation is now solved.

$$x=\frac{1-\sqrt{61}}{8}$$ $$x=\frac{\sqrt{61}+1}{8}$$

Steps for Completing the Square

Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $12x$, the least common multiple of $4x,3$.

$$3\times 5-4x\left(x+2\right)=12xx+12x\left(-1\right)$$

Multiply $3$ and $5$ to get $15$.

$$15-4x\left(x+2\right)=12xx+12x\left(-1\right)$$

Use the distributive property to multiply $4x$ by $x+2$.

$$15-\left(4x^{2}+8x\right)=12xx+12x\left(-1\right)$$

To find the opposite of $4x^{2}+8x$, find the opposite of each term.

$$15-4x^{2}-8x=12xx+12x\left(-1\right)$$

Multiply $x$ and $x$ to get $x^{2}$.

$$15-4x^{2}-8x=12x^{2}+12x\left(-1\right)$$

Multiply $12$ and $-1$ to get $-12$.

$$15-4x^{2}-8x=12x^{2}-12x$$

Subtract $12x^{2}$ from both sides.

$$15-4x^{2}-8x-12x^{2}=-12x$$

Combine $-4x^{2}$ and $-12x^{2}$ to get $-16x^{2}$.

$$15-16x^{2}-8x=-12x$$

Add $12x$ to both sides.

$$15-16x^{2}-8x+12x=0$$

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

$$15-16x^{2}+4x=0$$

Subtract $15$ from both sides. Anything subtracted from zero gives its negation.

$$-16x^{2}+4x=-15$$

Divide both sides by $-16$.

$$\frac{-16x^{2}+4x}{-16}=-\frac{15}{-16}$$

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

$$x^{2}+\frac{4}{-16}x=-\frac{15}{-16}$$

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

$$x^{2}-\frac{1}{4}x=-\frac{15}{-16}$$

Divide $-15$ by $-16$.

$$x^{2}-\frac{1}{4}x=\frac{15}{16}$$

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.

$$x^{2}-\frac{1}{4}x+\left(-\frac{1}{8}\right)^{2}=\frac{15}{16}+\left(-\frac{1}{8}\right)^{2}$$

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

$$x^{2}-\frac{1}{4}x+\frac{1}{64}=\frac{15}{16}+\frac{1}{64}$$

Add $\frac{15}{16}$ to $\frac{1}{64}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{4}x+\frac{1}{64}=\frac{61}{64}$$

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

$$\left(x-\frac{1}{8}\right)^{2}=\frac{61}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{8}\right)^{2}}=\sqrt{\frac{61}{64}}$$

Simplify.

$$x-\frac{1}{8}=\frac{\sqrt{61}}{8}$$ $$x-\frac{1}{8}=-\frac{\sqrt{61}}{8}$$

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

$$x=\frac{\sqrt{61}+1}{8}$$ $$x=\frac{1-\sqrt{61}}{8}$$