How to solve 4x(x - 3) = 8

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve 4x(x - 3) = 8 | Equatix

Question

$$4x(x-3)=8$$

Solve for x

$x = \frac{\sqrt{17} + 3}{2} \approx 3.561552813$
$x=\frac{3-\sqrt{17}}{2}\approx -0.561552813$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Use the distributive property to multiply $4x$ by $x-3$.

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

Subtract $8$ from both sides.

$$4x^{2}-12x-8=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $4$ for $a$, $-12$ for $b$, and $-8$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 4\left(-8\right)}}{2\times 4}$$

Square $-12$.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 4\left(-8\right)}}{2\times 4}$$

Multiply $-4$ times $4$.

$$x=\frac{-\left(-12\right)±\sqrt{144-16\left(-8\right)}}{2\times 4}$$

Multiply $-16$ times $-8$.

$$x=\frac{-\left(-12\right)±\sqrt{144+128}}{2\times 4}$$

Add $144$ to $128$.

$$x=\frac{-\left(-12\right)±\sqrt{272}}{2\times 4}$$

Take the square root of $272$.

$$x=\frac{-\left(-12\right)±4\sqrt{17}}{2\times 4}$$

The opposite of $-12$ is $12$.

$$x=\frac{12±4\sqrt{17}}{2\times 4}$$

Multiply $2$ times $4$.

$$x=\frac{12±4\sqrt{17}}{8}$$

Now solve the equation $x=\frac{12±4\sqrt{17}}{8}$ when $±$ is plus. Add $12$ to $4\sqrt{17}$.

$$x=\frac{4\sqrt{17}+12}{8}$$

Divide $12+4\sqrt{17}$ by $8$.

$$x=\frac{\sqrt{17}+3}{2}$$

Now solve the equation $x=\frac{12±4\sqrt{17}}{8}$ when $±$ is minus. Subtract $4\sqrt{17}$ from $12$.

$$x=\frac{12-4\sqrt{17}}{8}$$

Divide $12-4\sqrt{17}$ by $8$.

$$x=\frac{3-\sqrt{17}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{17}+3}{2}$$ $$x=\frac{3-\sqrt{17}}{2}$$

Steps for Completing the Square

Use the distributive property to multiply $4x$ by $x-3$.

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

Divide both sides by $4$.

$$\frac{4x^{2}-12x}{4}=\frac{8}{4}$$

Dividing by $4$ undoes the multiplication by $4$.

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

Divide $-12$ by $4$.

$$x^{2}-3x=\frac{8}{4}$$

Divide $8$ by $4$.

$$x^{2}-3x=2$$

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

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

Square $-\frac{3}{2}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=2+\frac{9}{4}$$

Add $2$ to $\frac{9}{4}$.

$$x^{2}-3x+\frac{9}{4}=\frac{17}{4}$$

Factor $x^{2}-3x+\frac{9}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{17}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{17}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{\sqrt{17}}{2}$$ $$x-\frac{3}{2}=-\frac{\sqrt{17}}{2}$$

Add $\frac{3}{2}$ to both sides of the equation.

$$x=\frac{\sqrt{17}+3}{2}$$ $$x=\frac{3-\sqrt{17}}{2}$$

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Solve for x

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Steps for Completing the Square