How to solve 9x ^ 2 = 12x - 49

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 9x ^ 2 = 12x - 49 | Equatix

Question

$$9x^{2}=12x-49$$

Solve for x (complex solution)

$x=\frac{2}{3}+\sqrt{5}i\approx 0.666666667+2.236067977i$
$x=-\sqrt{5}i+\frac{2}{3}\approx 0.666666667-2.236067977i$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Subtract $12x$ from both sides.

$$9x^{2}-12x=-49$$

Add $49$ to both sides.

$$9x^{2}-12x+49=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $9$ for $a$, $-12$ for $b$, and $49$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 9\times 49}}{2\times 9}$$

Square $-12$.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 9\times 49}}{2\times 9}$$

Multiply $-4$ times $9$.

$$x=\frac{-\left(-12\right)±\sqrt{144-36\times 49}}{2\times 9}$$

Multiply $-36$ times $49$.

$$x=\frac{-\left(-12\right)±\sqrt{144-1764}}{2\times 9}$$

Add $144$ to $-1764$.

$$x=\frac{-\left(-12\right)±\sqrt{-1620}}{2\times 9}$$

Take the square root of $-1620$.

$$x=\frac{-\left(-12\right)±18\sqrt{5}i}{2\times 9}$$

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

$$x=\frac{12±18\sqrt{5}i}{2\times 9}$$

Multiply $2$ times $9$.

$$x=\frac{12±18\sqrt{5}i}{18}$$

Now solve the equation $x=\frac{12±18\sqrt{5}i}{18}$ when $±$ is plus. Add $12$ to $18i\sqrt{5}$.

$$x=\frac{12+18\sqrt{5}i}{18}$$

Divide $12+18i\sqrt{5}$ by $18$.

$$x=\frac{2}{3}+\sqrt{5}i$$

Now solve the equation $x=\frac{12±18\sqrt{5}i}{18}$ when $±$ is minus. Subtract $18i\sqrt{5}$ from $12$.

$$x=\frac{-18\sqrt{5}i+12}{18}$$

Divide $12-18i\sqrt{5}$ by $18$.

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

The equation is now solved.

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

Steps for Completing the Square

Subtract $12x$ from both sides.

$$9x^{2}-12x=-49$$

Divide both sides by $9$.

$$\frac{9x^{2}-12x}{9}=-\frac{49}{9}$$

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

$$x^{2}+\left(-\frac{12}{9}\right)x=-\frac{49}{9}$$

Reduce the fraction $\frac{-12}{9}$ to lowest terms by extracting and canceling out $3$.

$$x^{2}-\frac{4}{3}x=-\frac{49}{9}$$

Divide $-\frac{4}{3}$, the coefficient of the $x$ term, by $2$ to get $-\frac{2}{3}$. Then add the square of $-\frac{2}{3}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{4}{3}x+\left(-\frac{2}{3}\right)^{2}=-\frac{49}{9}+\left(-\frac{2}{3}\right)^{2}$$

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

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{-49+4}{9}$$

Add $-\frac{49}{9}$ to $\frac{4}{9}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=-5$$

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

$$\left(x-\frac{2}{3}\right)^{2}=-5$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{3}\right)^{2}}=\sqrt{-5}$$

Simplify.

$$x-\frac{2}{3}=\sqrt{5}i$$ $$x-\frac{2}{3}=-\sqrt{5}i$$

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

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

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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