How to solve (x)^(2)+5x=-(3-x)

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 (x)^(2)+5x=-(3-x) | Equatix

Question

$${ x }^{ 2 } +5x=-(3-x)$$

Solve for x

$x=-3$
$x=-1$

Steps Using Factoring

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring

To find the opposite of $3-x$, find the opposite of each term.

$$x^{2}+5x=-3+x$$

Subtract $-3$ from both sides.

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

The opposite of $-3$ is $3$.

$$x^{2}+5x+3=x$$

Subtract $x$ from both sides.

$$x^{2}+5x+3-x=0$$

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

$$x^{2}+4x+3=0$$

To solve the equation, factor $x^{2}+4x+3$ using formula $x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right)$. To find $a$ and $b$, set up a system to be solved.

$$a+b=4$$ $$ab=3$$

Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. The only such pair is the system solution.

$$a=1$$ $$b=3$$

Rewrite factored expression $\left(x+a\right)\left(x+b\right)$ using the obtained values.

$$\left(x+1\right)\left(x+3\right)$$

To find equation solutions, solve $x+1=0$ and $x+3=0$.

$$x=-1$$ $$x=-3$$

Steps Using Factoring By Grouping

To find the opposite of $3-x$, find the opposite of each term.

$$x^{2}+5x=-3+x$$

Subtract $-3$ from both sides.

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

The opposite of $-3$ is $3$.

$$x^{2}+5x+3=x$$

Subtract $x$ from both sides.

$$x^{2}+5x+3-x=0$$

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

$$x^{2}+4x+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $x^{2}+ax+bx+3$. To find $a$ and $b$, set up a system to be solved.

$$a+b=4$$ $$ab=1\times 3=3$$

Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. The only such pair is the system solution.

$$a=1$$ $$b=3$$

Rewrite $x^{2}+4x+3$ as $\left(x^{2}+x\right)+\left(3x+3\right)$.

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

Factor out $x$ in the first and $3$ in the second group.

$$x\left(x+1\right)+3\left(x+1\right)$$

Factor out common term $x+1$ by using distributive property.

$$\left(x+1\right)\left(x+3\right)$$

To find equation solutions, solve $x+1=0$ and $x+3=0$.

$$x=-1$$ $$x=-3$$

Steps Using the Quadratic Formula

To find the opposite of $3-x$, find the opposite of each term.

$$x^{2}+5x=-3+x$$

Subtract $-3$ from both sides.

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

The opposite of $-3$ is $3$.

$$x^{2}+5x+3=x$$

Subtract $x$ from both sides.

$$x^{2}+5x+3-x=0$$

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

$$x^{2}+4x+3=0$$

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

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

Square $4$.

$$x=\frac{-4±\sqrt{16-4\times 3}}{2}$$

Multiply $-4$ times $3$.

$$x=\frac{-4±\sqrt{16-12}}{2}$$

Add $16$ to $-12$.

$$x=\frac{-4±\sqrt{4}}{2}$$

Take the square root of $4$.

$$x=\frac{-4±2}{2}$$

Now solve the equation $x=\frac{-4±2}{2}$ when $±$ is plus. Add $-4$ to $2$.

$$x=-\frac{2}{2}$$

Divide $-2$ by $2$.

$$x=-1$$

Now solve the equation $x=\frac{-4±2}{2}$ when $±$ is minus. Subtract $2$ from $-4$.

$$x=-\frac{6}{2}$$

Divide $-6$ by $2$.

$$x=-3$$

The equation is now solved.

$$x=-1$$ $$x=-3$$

Steps for Completing the Square

To find the opposite of $3-x$, find the opposite of each term.

$$x^{2}+5x=-3+x$$

Subtract $x$ from both sides.

$$x^{2}+5x-x=-3$$

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

$$x^{2}+4x=-3$$

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

$$x^{2}+4x+2^{2}=-3+2^{2}$$

Square $2$.

$$x^{2}+4x+4=-3+4$$

Add $-3$ to $4$.

$$x^{2}+4x+4=1$$

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

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{1}$$

Simplify.

$$x+2=1$$ $$x+2=-1$$

Subtract $2$ from both sides of the equation.

$$x=-1$$ $$x=-3$$

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Example

Question

Solve for x

Steps Using Factoring

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square