Solve x/2 - (2x)/3 * (3x)/4 = 5 | 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

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

Solve for x (complex solution)

$x=\frac{1+\sqrt{39}i}{2}\approx 0.5+3.122498999i$
$x=\frac{-\sqrt{39}i+1}{2}\approx 0.5-3.122498999i$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Multiply both sides of the equation by $12$, the least common multiple of $2,3,4$.

$$6x-2x\times 3x=60$$

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

$$6x-2x^{2}\times 3=60$$

Multiply $2$ and $3$ to get $6$.

$$6x-6x^{2}=60$$

Subtract $60$ from both sides.

$$6x-6x^{2}-60=0$$

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.

$$-6x^{2}+6x-60=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-6$ for $a$, $6$ for $b$, and $-60$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-6\right)\left(-60\right)}}{2\left(-6\right)}$$

Square $6$.

$$x=\frac{-6±\sqrt{36-4\left(-6\right)\left(-60\right)}}{2\left(-6\right)}$$

Multiply $-4$ times $-6$.

$$x=\frac{-6±\sqrt{36+24\left(-60\right)}}{2\left(-6\right)}$$

Multiply $24$ times $-60$.

$$x=\frac{-6±\sqrt{36-1440}}{2\left(-6\right)}$$

Add $36$ to $-1440$.

$$x=\frac{-6±\sqrt{-1404}}{2\left(-6\right)}$$

Take the square root of $-1404$.

$$x=\frac{-6±6\sqrt{39}i}{2\left(-6\right)}$$

Multiply $2$ times $-6$.

$$x=\frac{-6±6\sqrt{39}i}{-12}$$

Now solve the equation $x=\frac{-6±6\sqrt{39}i}{-12}$ when $±$ is plus. Add $-6$ to $6i\sqrt{39}$.

$$x=\frac{-6+6\sqrt{39}i}{-12}$$

Divide $-6+6i\sqrt{39}$ by $-12$.

$$x=\frac{-\sqrt{39}i+1}{2}$$

Now solve the equation $x=\frac{-6±6\sqrt{39}i}{-12}$ when $±$ is minus. Subtract $6i\sqrt{39}$ from $-6$.

$$x=\frac{-6\sqrt{39}i-6}{-12}$$

Divide $-6-6i\sqrt{39}$ by $-12$.

$$x=\frac{1+\sqrt{39}i}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{39}i+1}{2}$$ $$x=\frac{1+\sqrt{39}i}{2}$$

Steps for Completing the Square

Multiply both sides of the equation by $12$, the least common multiple of $2,3,4$.

$$6x-2x\times 3x=60$$

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

$$6x-2x^{2}\times 3=60$$

Multiply $2$ and $3$ to get $6$.

$$6x-6x^{2}=60$$

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 $x^{2}+bx=c$.

$$-6x^{2}+6x=60$$

Divide both sides by $-6$.

$$\frac{-6x^{2}+6x}{-6}=\frac{60}{-6}$$

Dividing by $-6$ undoes the multiplication by $-6$.

$$x^{2}+\frac{6}{-6}x=\frac{60}{-6}$$

Divide $6$ by $-6$.

$$x^{2}-x=\frac{60}{-6}$$

Divide $60$ by $-6$.

$$x^{2}-x=-10$$

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

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=-10+\left(-\frac{1}{2}\right)^{2}$$

Square $-\frac{1}{2}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=-10+\frac{1}{4}$$

Add $-10$ to $\frac{1}{4}$.

$$x^{2}-x+\frac{1}{4}=-\frac{39}{4}$$

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

$$\left(x-\frac{1}{2}\right)^{2}=-\frac{39}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{-\frac{39}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{39}i}{2}$$ $$x-\frac{1}{2}=-\frac{\sqrt{39}i}{2}$$

Add $\frac{1}{2}$ to both sides of the equation.

$$x=\frac{1+\sqrt{39}i}{2}$$ $$x=\frac{-\sqrt{39}i+1}{2}$$