Solve 11r + 15 = - 2r ^ 2 | 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

$$11r+15=-2r^{2}$$

Solve for r

$r = -\frac{5}{2} = -2\frac{1}{2} = -2.5$
$r=-3$

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring By Grouping

Add $2r^{2}$ to both sides.

$$11r+15+2r^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$2r^{2}+11r+15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $2r^{2}+ar+br+15$. To find $a$ and $b$, set up a system to be solved.

$$a+b=11$$ $$ab=2\times 15=30$$

Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. List all such integer pairs that give product $30$.

$$1,30$$ $$2,15$$ $$3,10$$ $$5,6$$

Calculate the sum for each pair.

$$1+30=31$$ $$2+15=17$$ $$3+10=13$$ $$5+6=11$$

The solution is the pair that gives sum $11$.

$$a=5$$ $$b=6$$

Rewrite $2r^{2}+11r+15$ as $\left(2r^{2}+5r\right)+\left(6r+15\right)$.

$$\left(2r^{2}+5r\right)+\left(6r+15\right)$$

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

$$r\left(2r+5\right)+3\left(2r+5\right)$$

Factor out common term $2r+5$ by using distributive property.

$$\left(2r+5\right)\left(r+3\right)$$

To find equation solutions, solve $2r+5=0$ and $r+3=0$.

$$r=-\frac{5}{2}$$ $$r=-3$$

Steps Using the Quadratic Formula

Add $2r^{2}$ to both sides.

$$11r+15+2r^{2}=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.

$$2r^{2}+11r+15=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $2$ for $a$, $11$ for $b$, and $15$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$r=\frac{-11±\sqrt{11^{2}-4\times 2\times 15}}{2\times 2}$$

Square $11$.

$$r=\frac{-11±\sqrt{121-4\times 2\times 15}}{2\times 2}$$

Multiply $-4$ times $2$.

$$r=\frac{-11±\sqrt{121-8\times 15}}{2\times 2}$$

Multiply $-8$ times $15$.

$$r=\frac{-11±\sqrt{121-120}}{2\times 2}$$

Add $121$ to $-120$.

$$r=\frac{-11±\sqrt{1}}{2\times 2}$$

Take the square root of $1$.

$$r=\frac{-11±1}{2\times 2}$$

Multiply $2$ times $2$.

$$r=\frac{-11±1}{4}$$

Now solve the equation $r=\frac{-11±1}{4}$ when $±$ is plus. Add $-11$ to $1$.

$$r=-\frac{10}{4}$$

Reduce the fraction $\frac{-10}{4}$ to lowest terms by extracting and canceling out $2$.

$$r=-\frac{5}{2}$$

Now solve the equation $r=\frac{-11±1}{4}$ when $±$ is minus. Subtract $1$ from $-11$.

$$r=-\frac{12}{4}$$

Divide $-12$ by $4$.

$$r=-3$$

The equation is now solved.

$$r=-\frac{5}{2}$$ $$r=-3$$

Steps for Completing the Square

Add $2r^{2}$ to both sides.

$$11r+15+2r^{2}=0$$

Subtract $15$ from both sides. Anything subtracted from zero gives its negation.

$$11r+2r^{2}=-15$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.

$$2r^{2}+11r=-15$$

Divide both sides by $2$.

$$\frac{2r^{2}+11r}{2}=-\frac{15}{2}$$

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

$$r^{2}+\frac{11}{2}r=-\frac{15}{2}$$

Divide $\frac{11}{2}$, the coefficient of the $x$ term, by $2$ to get $\frac{11}{4}$. Then add the square of $\frac{11}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$r^{2}+\frac{11}{2}r+\left(\frac{11}{4}\right)^{2}=-\frac{15}{2}+\left(\frac{11}{4}\right)^{2}$$

Square $\frac{11}{4}$ by squaring both the numerator and the denominator of the fraction.

$$r^{2}+\frac{11}{2}r+\frac{121}{16}=-\frac{15}{2}+\frac{121}{16}$$

Add $-\frac{15}{2}$ to $\frac{121}{16}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$r^{2}+\frac{11}{2}r+\frac{121}{16}=\frac{1}{16}$$

Factor $r^{2}+\frac{11}{2}r+\frac{121}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(r+\frac{11}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(r+\frac{11}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$r+\frac{11}{4}=\frac{1}{4}$$ $$r+\frac{11}{4}=-\frac{1}{4}$$

Subtract $\frac{11}{4}$ from both sides of the equation.

$$r=-\frac{5}{2}$$ $$r=-3$$