How to solve 3p(11 - 2p) = 15

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 3p(11 - 2p) = 15 | Equatix

Question

$$3p(11-2p)=15$$

Solve for p

$p=5$
$p=\frac{1}{2}=0.5$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Use the distributive property to multiply $3p$ by $11-2p$.

$$33p-6p^{2}=15$$

Subtract $15$ from both sides.

$$33p-6p^{2}-15=0$$

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.

$$-6p^{2}+33p-15=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-6$ for $a$, $33$ for $b$, and $-15$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$p=\frac{-33±\sqrt{33^{2}-4\left(-6\right)\left(-15\right)}}{2\left(-6\right)}$$

Square $33$.

$$p=\frac{-33±\sqrt{1089-4\left(-6\right)\left(-15\right)}}{2\left(-6\right)}$$

Multiply $-4$ times $-6$.

$$p=\frac{-33±\sqrt{1089+24\left(-15\right)}}{2\left(-6\right)}$$

Multiply $24$ times $-15$.

$$p=\frac{-33±\sqrt{1089-360}}{2\left(-6\right)}$$

Add $1089$ to $-360$.

$$p=\frac{-33±\sqrt{729}}{2\left(-6\right)}$$

Take the square root of $729$.

$$p=\frac{-33±27}{2\left(-6\right)}$$

Multiply $2$ times $-6$.

$$p=\frac{-33±27}{-12}$$

Now solve the equation $p=\frac{-33±27}{-12}$ when $±$ is plus. Add $-33$ to $27$.

$$p=-\frac{6}{-12}$$

Reduce the fraction $\frac{-6}{-12}$ to lowest terms by extracting and canceling out $6$.

$$p=\frac{1}{2}$$

Now solve the equation $p=\frac{-33±27}{-12}$ when $±$ is minus. Subtract $27$ from $-33$.

$$p=-\frac{60}{-12}$$

Divide $-60$ by $-12$.

$$p=5$$

The equation is now solved.

$$p=\frac{1}{2}$$ $$p=5$$

Steps for Completing the Square

Use the distributive property to multiply $3p$ by $11-2p$.

$$33p-6p^{2}=15$$

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 $x^{2}+bx=c$.

$$-6p^{2}+33p=15$$

Divide both sides by $-6$.

$$\frac{-6p^{2}+33p}{-6}=\frac{15}{-6}$$

Dividing by $-6$ undoes the multiplication by $-6$.

$$p^{2}+\frac{33}{-6}p=\frac{15}{-6}$$

Reduce the fraction $\frac{33}{-6}$ to lowest terms by extracting and canceling out $3$.

$$p^{2}-\frac{11}{2}p=\frac{15}{-6}$$

Reduce the fraction $\frac{15}{-6}$ to lowest terms by extracting and canceling out $3$.

$$p^{2}-\frac{11}{2}p=-\frac{5}{2}$$

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.

$$p^{2}-\frac{11}{2}p+\left(-\frac{11}{4}\right)^{2}=-\frac{5}{2}+\left(-\frac{11}{4}\right)^{2}$$

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

$$p^{2}-\frac{11}{2}p+\frac{121}{16}=-\frac{5}{2}+\frac{121}{16}$$

Add $-\frac{5}{2}$ to $\frac{121}{16}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$p^{2}-\frac{11}{2}p+\frac{121}{16}=\frac{81}{16}$$

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

$$\left(p-\frac{11}{4}\right)^{2}=\frac{81}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p-\frac{11}{4}\right)^{2}}=\sqrt{\frac{81}{16}}$$

Simplify.

$$p-\frac{11}{4}=\frac{9}{4}$$ $$p-\frac{11}{4}=-\frac{9}{4}$$

Add $\frac{11}{4}$ to both sides of the equation.

$$p=5$$ $$p=\frac{1}{2}$$

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Solve for p

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Steps for Completing the Square