How to solve 9(x)^(2)+8=6x

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 9(x)^(2)+8=6x | Equatix

Question

$$9 { x }^{ 2 } +8=6x$$

Solve for x (complex solution)

$x=\frac{1+\sqrt{7}i}{3}\approx 0.333333333+0.881917104i$
$x=\frac{-\sqrt{7}i+1}{3}\approx 0.333333333-0.881917104i$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Subtract $6x$ from both sides.

$$9x^{2}+8-6x=0$$

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.

$$9x^{2}-6x+8=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $9$ for $a$, $-6$ for $b$, and $8$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 9\times 8}}{2\times 9}$$

Square $-6$.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 9\times 8}}{2\times 9}$$

Multiply $-4$ times $9$.

$$x=\frac{-\left(-6\right)±\sqrt{36-36\times 8}}{2\times 9}$$

Multiply $-36$ times $8$.

$$x=\frac{-\left(-6\right)±\sqrt{36-288}}{2\times 9}$$

Add $36$ to $-288$.

$$x=\frac{-\left(-6\right)±\sqrt{-252}}{2\times 9}$$

Take the square root of $-252$.

$$x=\frac{-\left(-6\right)±6\sqrt{7}i}{2\times 9}$$

The opposite of $-6$ is $6$.

$$x=\frac{6±6\sqrt{7}i}{2\times 9}$$

Multiply $2$ times $9$.

$$x=\frac{6±6\sqrt{7}i}{18}$$

Now solve the equation $x=\frac{6±6\sqrt{7}i}{18}$ when $±$ is plus. Add $6$ to $6i\sqrt{7}$.

$$x=\frac{6+6\sqrt{7}i}{18}$$

Divide $6+6i\sqrt{7}$ by $18$.

$$x=\frac{1+\sqrt{7}i}{3}$$

Now solve the equation $x=\frac{6±6\sqrt{7}i}{18}$ when $±$ is minus. Subtract $6i\sqrt{7}$ from $6$.

$$x=\frac{-6\sqrt{7}i+6}{18}$$

Divide $6-6i\sqrt{7}$ by $18$.

$$x=\frac{-\sqrt{7}i+1}{3}$$

The equation is now solved.

$$x=\frac{1+\sqrt{7}i}{3}$$ $$x=\frac{-\sqrt{7}i+1}{3}$$

Steps for Completing the Square

Subtract $6x$ from both sides.

$$9x^{2}+8-6x=0$$

Subtract $8$ from both sides. Anything subtracted from zero gives its negation.

$$9x^{2}-6x=-8$$

Divide both sides by $9$.

$$\frac{9x^{2}-6x}{9}=-\frac{8}{9}$$

Dividing by $9$ undoes the multiplication by $9$.

$$x^{2}+\left(-\frac{6}{9}\right)x=-\frac{8}{9}$$

Reduce the fraction $\frac{-6}{9}$ to lowest terms by extracting and canceling out $3$.

$$x^{2}-\frac{2}{3}x=-\frac{8}{9}$$

Divide $-\frac{2}{3}$, the coefficient of the $x$ term, by $2$ to get $-\frac{1}{3}$. Then add the square of $-\frac{1}{3}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{2}{3}x+\left(-\frac{1}{3}\right)^{2}=-\frac{8}{9}+\left(-\frac{1}{3}\right)^{2}$$

Square $-\frac{1}{3}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{-8+1}{9}$$

Add $-\frac{8}{9}$ to $\frac{1}{9}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=-\frac{7}{9}$$

Factor $x^{2}-\frac{2}{3}x+\frac{1}{9}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{1}{3}\right)^{2}=-\frac{7}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{-\frac{7}{9}}$$

Simplify.

$$x-\frac{1}{3}=\frac{\sqrt{7}i}{3}$$ $$x-\frac{1}{3}=-\frac{\sqrt{7}i}{3}$$

Add $\frac{1}{3}$ to both sides of the equation.

$$x=\frac{1+\sqrt{7}i}{3}$$ $$x=\frac{-\sqrt{7}i+1}{3}$$

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Example

Question

Solve for x (complex solution)

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Steps for Completing the Square