Solve x(x+4)+(x+2)=6(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

$$x(x+4)+(x+2)=6(x+3)$$

Solve for x

$x = \frac{\sqrt{65} + 1}{2} \approx 4.531128874$
$x=\frac{1-\sqrt{65}}{2}\approx -3.531128874$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

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

$$x^{2}+4x+x+2=6\left(x+3\right)$$

Combine $4x$ and $x$ to get $5x$.

$$x^{2}+5x+2=6\left(x+3\right)$$

Use the distributive property to multiply $6$ by $x+3$.

$$x^{2}+5x+2=6x+18$$

Subtract $6x$ from both sides.

$$x^{2}+5x+2-6x=18$$

Combine $5x$ and $-6x$ to get $-x$.

$$x^{2}-x+2=18$$

Subtract $18$ from both sides.

$$x^{2}-x+2-18=0$$

Subtract $18$ from $2$ to get $-16$.

$$x^{2}-x-16=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $-1$ for $b$, and $-16$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-16\right)}}{2}$$

Multiply $-4$ times $-16$.

$$x=\frac{-\left(-1\right)±\sqrt{1+64}}{2}$$

Add $1$ to $64$.

$$x=\frac{-\left(-1\right)±\sqrt{65}}{2}$$

The opposite of $-1$ is $1$.

$$x=\frac{1±\sqrt{65}}{2}$$

Now solve the equation $x=\frac{1±\sqrt{65}}{2}$ when $±$ is plus. Add $1$ to $\sqrt{65}$.

$$x=\frac{\sqrt{65}+1}{2}$$

Now solve the equation $x=\frac{1±\sqrt{65}}{2}$ when $±$ is minus. Subtract $\sqrt{65}$ from $1$.

$$x=\frac{1-\sqrt{65}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{65}+1}{2}$$ $$x=\frac{1-\sqrt{65}}{2}$$

Steps for Completing the Square

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

$$x^{2}+4x+x+2=6\left(x+3\right)$$

Combine $4x$ and $x$ to get $5x$.

$$x^{2}+5x+2=6\left(x+3\right)$$

Use the distributive property to multiply $6$ by $x+3$.

$$x^{2}+5x+2=6x+18$$

Subtract $6x$ from both sides.

$$x^{2}+5x+2-6x=18$$

Combine $5x$ and $-6x$ to get $-x$.

$$x^{2}-x+2=18$$

Subtract $2$ from both sides.

$$x^{2}-x=18-2$$

Subtract $2$ from $18$ to get $16$.

$$x^{2}-x=16$$

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

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=16+\left(-\frac{1}{2}\right)^{2}$$

Square $-\frac{1}{2}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=16+\frac{1}{4}$$

Add $16$ to $\frac{1}{4}$.

$$x^{2}-x+\frac{1}{4}=\frac{65}{4}$$

Factor $x^{2}-x+\frac{1}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{65}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{65}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{65}}{2}$$ $$x-\frac{1}{2}=-\frac{\sqrt{65}}{2}$$

Add $\frac{1}{2}$ to both sides of the equation.

$$x=\frac{\sqrt{65}+1}{2}$$ $$x=\frac{1-\sqrt{65}}{2}$$