How to solve + + = [ + 4

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 + + = [ + 4 | Equatix

Question

$$( x ^ { 2 } + 4 x ) + 81 = [ ( x + 4 ) 3 ] ( x + 3 )$$

Solve for x

$x = \frac{\sqrt{649} - 17}{4} \approx 2.118869601$
$x=\frac{-\sqrt{649}-17}{4}\approx -10.618869601$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

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

$$x^{2}+4x+81=\left(3x+12\right)\left(x+3\right)$$

Use the distributive property to multiply $3x+12$ by $x+3$ and combine like terms.

$$x^{2}+4x+81=3x^{2}+21x+36$$

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

$$x^{2}+4x+81-3x^{2}=21x+36$$

Combine $x^{2}$ and $-3x^{2}$ to get $-2x^{2}$.

$$-2x^{2}+4x+81=21x+36$$

Subtract $21x$ from both sides.

$$-2x^{2}+4x+81-21x=36$$

Combine $4x$ and $-21x$ to get $-17x$.

$$-2x^{2}-17x+81=36$$

Subtract $36$ from both sides.

$$-2x^{2}-17x+81-36=0$$

Subtract $36$ from $81$ to get $45$.

$$-2x^{2}-17x+45=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-2$ for $a$, $-17$ for $b$, and $45$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\left(-2\right)\times 45}}{2\left(-2\right)}$$

Square $-17$.

$$x=\frac{-\left(-17\right)±\sqrt{289-4\left(-2\right)\times 45}}{2\left(-2\right)}$$

Multiply $-4$ times $-2$.

$$x=\frac{-\left(-17\right)±\sqrt{289+8\times 45}}{2\left(-2\right)}$$

Multiply $8$ times $45$.

$$x=\frac{-\left(-17\right)±\sqrt{289+360}}{2\left(-2\right)}$$

Add $289$ to $360$.

$$x=\frac{-\left(-17\right)±\sqrt{649}}{2\left(-2\right)}$$

The opposite of $-17$ is $17$.

$$x=\frac{17±\sqrt{649}}{2\left(-2\right)}$$

Multiply $2$ times $-2$.

$$x=\frac{17±\sqrt{649}}{-4}$$

Now solve the equation $x=\frac{17±\sqrt{649}}{-4}$ when $±$ is plus. Add $17$ to $\sqrt{649}$.

$$x=\frac{\sqrt{649}+17}{-4}$$

Divide $17+\sqrt{649}$ by $-4$.

$$x=\frac{-\sqrt{649}-17}{4}$$

Now solve the equation $x=\frac{17±\sqrt{649}}{-4}$ when $±$ is minus. Subtract $\sqrt{649}$ from $17$.

$$x=\frac{17-\sqrt{649}}{-4}$$

Divide $17-\sqrt{649}$ by $-4$.

$$x=\frac{\sqrt{649}-17}{4}$$

The equation is now solved.

$$x=\frac{-\sqrt{649}-17}{4}$$ $$x=\frac{\sqrt{649}-17}{4}$$

Steps for Completing the Square

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

$$x^{2}+4x+81=\left(3x+12\right)\left(x+3\right)$$

Use the distributive property to multiply $3x+12$ by $x+3$ and combine like terms.

$$x^{2}+4x+81=3x^{2}+21x+36$$

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

$$x^{2}+4x+81-3x^{2}=21x+36$$

Combine $x^{2}$ and $-3x^{2}$ to get $-2x^{2}$.

$$-2x^{2}+4x+81=21x+36$$

Subtract $21x$ from both sides.

$$-2x^{2}+4x+81-21x=36$$

Combine $4x$ and $-21x$ to get $-17x$.

$$-2x^{2}-17x+81=36$$

Subtract $81$ from both sides.

$$-2x^{2}-17x=36-81$$

Subtract $81$ from $36$ to get $-45$.

$$-2x^{2}-17x=-45$$

Divide both sides by $-2$.

$$\frac{-2x^{2}-17x}{-2}=-\frac{45}{-2}$$

Dividing by $-2$ undoes the multiplication by $-2$.

$$x^{2}+\left(-\frac{17}{-2}\right)x=-\frac{45}{-2}$$

Divide $-17$ by $-2$.

$$x^{2}+\frac{17}{2}x=-\frac{45}{-2}$$

Divide $-45$ by $-2$.

$$x^{2}+\frac{17}{2}x=\frac{45}{2}$$

Divide $\frac{17}{2}$, the coefficient of the $x$ term, by $2$ to get $\frac{17}{4}$. Then add the square of $\frac{17}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{17}{2}x+\left(\frac{17}{4}\right)^{2}=\frac{45}{2}+\left(\frac{17}{4}\right)^{2}$$

Square $\frac{17}{4}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{17}{2}x+\frac{289}{16}=\frac{45}{2}+\frac{289}{16}$$

Add $\frac{45}{2}$ to $\frac{289}{16}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{17}{2}x+\frac{289}{16}=\frac{649}{16}$$

Factor $x^{2}+\frac{17}{2}x+\frac{289}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{17}{4}\right)^{2}=\frac{649}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{17}{4}\right)^{2}}=\sqrt{\frac{649}{16}}$$

Simplify.

$$x+\frac{17}{4}=\frac{\sqrt{649}}{4}$$ $$x+\frac{17}{4}=-\frac{\sqrt{649}}{4}$$

Subtract $\frac{17}{4}$ from both sides of the equation.

$$x=\frac{\sqrt{649}-17}{4}$$ $$x=\frac{-\sqrt{649}-17}{4}$$

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Solve for x

Steps Using the Quadratic Formula

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Steps Using the Quadratic Formula

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