Solve (x^ 2 +4x)+81=[x+4)+3](x+3) | 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

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

Solve for x (complex solution)

$x=\frac{-3+\sqrt{287}i}{2}\approx -1.5+8.470537173i$
$x=\frac{-\sqrt{287}i-3}{2}\approx -1.5-8.470537173i$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Add $4$ and $3$ to get $7$.

$$x^{2}+4x+81=x+7$$

Subtract $x$ from both sides.

$$x^{2}+4x+81-x=7$$

Combine $4x$ and $-x$ to get $3x$.

$$x^{2}+3x+81=7$$

Subtract $7$ from both sides.

$$x^{2}+3x+81-7=0$$

Subtract $7$ from $81$ to get $74$.

$$x^{2}+3x+74=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $3$ for $b$, and $74$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-3±\sqrt{3^{2}-4\times 74}}{2}$$

Square $3$.

$$x=\frac{-3±\sqrt{9-4\times 74}}{2}$$

Multiply $-4$ times $74$.

$$x=\frac{-3±\sqrt{9-296}}{2}$$

Add $9$ to $-296$.

$$x=\frac{-3±\sqrt{-287}}{2}$$

Take the square root of $-287$.

$$x=\frac{-3±\sqrt{287}i}{2}$$

Now solve the equation $x=\frac{-3±\sqrt{287}i}{2}$ when $±$ is plus. Add $-3$ to $i\sqrt{287}$.

$$x=\frac{-3+\sqrt{287}i}{2}$$

Now solve the equation $x=\frac{-3±\sqrt{287}i}{2}$ when $±$ is minus. Subtract $i\sqrt{287}$ from $-3$.

$$x=\frac{-\sqrt{287}i-3}{2}$$

The equation is now solved.

$$x=\frac{-3+\sqrt{287}i}{2}$$ $$x=\frac{-\sqrt{287}i-3}{2}$$

Steps for Completing the Square

Add $4$ and $3$ to get $7$.

$$x^{2}+4x+81=x+7$$

Subtract $x$ from both sides.

$$x^{2}+4x+81-x=7$$

Combine $4x$ and $-x$ to get $3x$.

$$x^{2}+3x+81=7$$

Subtract $81$ from both sides.

$$x^{2}+3x=7-81$$

Subtract $81$ from $7$ to get $-74$.

$$x^{2}+3x=-74$$

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.

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=-74+\left(\frac{3}{2}\right)^{2}$$

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

$$x^{2}+3x+\frac{9}{4}=-74+\frac{9}{4}$$

Add $-74$ to $\frac{9}{4}$.

$$x^{2}+3x+\frac{9}{4}=-\frac{287}{4}$$

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

$$\left(x+\frac{3}{2}\right)^{2}=-\frac{287}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{-\frac{287}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{\sqrt{287}i}{2}$$ $$x+\frac{3}{2}=-\frac{\sqrt{287}i}{2}$$

Subtract $\frac{3}{2}$ from both sides of the equation.

$$x=\frac{-3+\sqrt{287}i}{2}$$ $$x=\frac{-\sqrt{287}i-3}{2}$$