Solve (sqrt(3))/(4)\times4x+4=(sqrt(3)(x)^(2)+24sqrt(3))/(4) | 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

$$\frac{ \sqrt{ 3 } }{ 4 } \times 4x+4= \frac{ \sqrt{ 3 } { x }^{ 2 } +24 \sqrt{ 3 } }{ 4 }$$

Solve for x (complex solution)

$x=-\frac{2i\sqrt{45-12\sqrt{3}}}{3}+2\approx 2-3.280609043i$
$x=\frac{2i\sqrt{45-12\sqrt{3}}}{3}+2\approx 2+3.280609043i$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Multiply both sides of the equation by $4$.

$$4\sqrt{3}x+16=\sqrt{3}x^{2}+24\sqrt{3}$$

Subtract $\sqrt{3}x^{2}$ from both sides.

$$4\sqrt{3}x+16-\sqrt{3}x^{2}=24\sqrt{3}$$

Subtract $24\sqrt{3}$ from both sides.

$$4\sqrt{3}x+16-\sqrt{3}x^{2}-24\sqrt{3}=0$$

Reorder the terms.

$$-\sqrt{3}x^{2}+4\sqrt{3}x-24\sqrt{3}+16=0$$

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.

$$\left(-\sqrt{3}\right)x^{2}+4\sqrt{3}x+16-24\sqrt{3}=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-\sqrt{3}$ for $a$, $4\sqrt{3}$ for $b$, and $-24\sqrt{3}+16$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-4\sqrt{3}±\sqrt{\left(4\sqrt{3}\right)^{2}-4\left(-\sqrt{3}\right)\left(16-24\sqrt{3}\right)}}{2\left(-\sqrt{3}\right)}$$

Square $4\sqrt{3}$.

$$x=\frac{-4\sqrt{3}±\sqrt{48-4\left(-\sqrt{3}\right)\left(16-24\sqrt{3}\right)}}{2\left(-\sqrt{3}\right)}$$

Multiply $-4$ times $-\sqrt{3}$.

$$x=\frac{-4\sqrt{3}±\sqrt{48+4\sqrt{3}\left(16-24\sqrt{3}\right)}}{2\left(-\sqrt{3}\right)}$$

Multiply $4\sqrt{3}$ times $-24\sqrt{3}+16$.

$$x=\frac{-4\sqrt{3}±\sqrt{48+64\sqrt{3}-288}}{2\left(-\sqrt{3}\right)}$$

Add $48$ to $-288+64\sqrt{3}$.

$$x=\frac{-4\sqrt{3}±\sqrt{64\sqrt{3}-240}}{2\left(-\sqrt{3}\right)}$$

Take the square root of $-240+64\sqrt{3}$.

$$x=\frac{-4\sqrt{3}±4i\sqrt{15-4\sqrt{3}}}{2\left(-\sqrt{3}\right)}$$

Multiply $2$ times $-\sqrt{3}$.

$$x=\frac{-4\sqrt{3}±4i\sqrt{15-4\sqrt{3}}}{-2\sqrt{3}}$$

Now solve the equation $x=\frac{-4\sqrt{3}±4i\sqrt{15-4\sqrt{3}}}{-2\sqrt{3}}$ when $±$ is plus. Add $-4\sqrt{3}$ to $4i\sqrt{15-4\sqrt{3}}$.

$$x=\frac{-4\sqrt{3}+4i\sqrt{15-4\sqrt{3}}}{-2\sqrt{3}}$$

Divide $-4\sqrt{3}+4i\sqrt{15-4\sqrt{3}}$ by $-2\sqrt{3}$.

$$x=-\frac{2i\sqrt{45-12\sqrt{3}}}{3}+2$$

Now solve the equation $x=\frac{-4\sqrt{3}±4i\sqrt{15-4\sqrt{3}}}{-2\sqrt{3}}$ when $±$ is minus. Subtract $4i\sqrt{15-4\sqrt{3}}$ from $-4\sqrt{3}$.

$$x=\frac{-4i\sqrt{15-4\sqrt{3}}-4\sqrt{3}}{-2\sqrt{3}}$$

Divide $-4\sqrt{3}-4i\sqrt{15-4\sqrt{3}}$ by $-2\sqrt{3}$.

$$x=\frac{2i\sqrt{45-12\sqrt{3}}}{3}+2$$

The equation is now solved.

$$x=-\frac{2i\sqrt{45-12\sqrt{3}}}{3}+2$$ $$x=\frac{2i\sqrt{45-12\sqrt{3}}}{3}+2$$

Steps for Completing the Square

Multiply both sides of the equation by $4$.

$$4\sqrt{3}x+16=\sqrt{3}x^{2}+24\sqrt{3}$$

Subtract $\sqrt{3}x^{2}$ from both sides.

$$4\sqrt{3}x+16-\sqrt{3}x^{2}=24\sqrt{3}$$

Subtract $16$ from both sides.

$$4\sqrt{3}x-\sqrt{3}x^{2}=24\sqrt{3}-16$$

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

$$\left(-\sqrt{3}\right)x^{2}+4\sqrt{3}x=24\sqrt{3}-16$$

Divide both sides by $-\sqrt{3}$.

$$\frac{\left(-\sqrt{3}\right)x^{2}+4\sqrt{3}x}{-\sqrt{3}}=\frac{24\sqrt{3}-16}{-\sqrt{3}}$$

Dividing by $-\sqrt{3}$ undoes the multiplication by $-\sqrt{3}$.

$$x^{2}+\frac{4\sqrt{3}}{-\sqrt{3}}x=\frac{24\sqrt{3}-16}{-\sqrt{3}}$$

Divide $4\sqrt{3}$ by $-\sqrt{3}$.

$$x^{2}-4x=\frac{24\sqrt{3}-16}{-\sqrt{3}}$$

Divide $24\sqrt{3}-16$ by $-\sqrt{3}$.

$$x^{2}-4x=\frac{16\sqrt{3}}{3}-24$$

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

$$x^{2}-4x+\left(-2\right)^{2}=\frac{16\sqrt{3}}{3}-24+\left(-2\right)^{2}$$

Square $-2$.

$$x^{2}-4x+4=\frac{16\sqrt{3}}{3}-24+4$$

Add $-24+\frac{16\sqrt{3}}{3}$ to $4$.

$$x^{2}-4x+4=\frac{16\sqrt{3}}{3}-20$$

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

$$\left(x-2\right)^{2}=\frac{16\sqrt{3}}{3}-20$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{\frac{16\sqrt{3}}{3}-20}$$

Simplify.

$$x-2=\frac{2i\sqrt{45-12\sqrt{3}}}{3}$$ $$x-2=-\frac{2i\sqrt{45-12\sqrt{3}}}{3}$$

Add $2$ to both sides of the equation.

$$x=\frac{2i\sqrt{45-12\sqrt{3}}}{3}+2$$ $$x=-\frac{2i\sqrt{45-12\sqrt{3}}}{3}+2$$