Solve 19 - 3P - 3P ^ 2 = 5P - 1 T | 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

$$19 - 3 p - 3 p ^ { 2 } = 5 p - 1$$

Solve for p

$p = \frac{2 \sqrt{19} - 4}{3} \approx 1.572599296$
$p=\frac{-2\sqrt{19}-4}{3}\approx -4.239265962$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Subtract $5p$ from both sides.

$$19-3p-3p^{2}-5p=-1$$

Combine $-3p$ and $-5p$ to get $-8p$.

$$19-8p-3p^{2}=-1$$

Add $1$ to both sides.

$$19-8p-3p^{2}+1=0$$

Add $19$ and $1$ to get $20$.

$$20-8p-3p^{2}=0$$

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.

$$-3p^{2}-8p+20=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-3$ for $a$, $-8$ for $b$, and $20$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$p=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-3\right)\times 20}}{2\left(-3\right)}$$

Square $-8$.

$$p=\frac{-\left(-8\right)±\sqrt{64-4\left(-3\right)\times 20}}{2\left(-3\right)}$$

Multiply $-4$ times $-3$.

$$p=\frac{-\left(-8\right)±\sqrt{64+12\times 20}}{2\left(-3\right)}$$

Multiply $12$ times $20$.

$$p=\frac{-\left(-8\right)±\sqrt{64+240}}{2\left(-3\right)}$$

Add $64$ to $240$.

$$p=\frac{-\left(-8\right)±\sqrt{304}}{2\left(-3\right)}$$

Take the square root of $304$.

$$p=\frac{-\left(-8\right)±4\sqrt{19}}{2\left(-3\right)}$$

The opposite of $-8$ is $8$.

$$p=\frac{8±4\sqrt{19}}{2\left(-3\right)}$$

Multiply $2$ times $-3$.

$$p=\frac{8±4\sqrt{19}}{-6}$$

Now solve the equation $p=\frac{8±4\sqrt{19}}{-6}$ when $±$ is plus. Add $8$ to $4\sqrt{19}$.

$$p=\frac{4\sqrt{19}+8}{-6}$$

Divide $8+4\sqrt{19}$ by $-6$.

$$p=\frac{-2\sqrt{19}-4}{3}$$

Now solve the equation $p=\frac{8±4\sqrt{19}}{-6}$ when $±$ is minus. Subtract $4\sqrt{19}$ from $8$.

$$p=\frac{8-4\sqrt{19}}{-6}$$

Divide $8-4\sqrt{19}$ by $-6$.

$$p=\frac{2\sqrt{19}-4}{3}$$

The equation is now solved.

$$p=\frac{-2\sqrt{19}-4}{3}$$ $$p=\frac{2\sqrt{19}-4}{3}$$

Steps for Completing the Square

Subtract $5p$ from both sides.

$$19-3p-3p^{2}-5p=-1$$

Combine $-3p$ and $-5p$ to get $-8p$.

$$19-8p-3p^{2}=-1$$

Subtract $19$ from both sides.

$$-8p-3p^{2}=-1-19$$

Subtract $19$ from $-1$ to get $-20$.

$$-8p-3p^{2}=-20$$

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

$$-3p^{2}-8p=-20$$

Divide both sides by $-3$.

$$\frac{-3p^{2}-8p}{-3}=-\frac{20}{-3}$$

Dividing by $-3$ undoes the multiplication by $-3$.

$$p^{2}+\left(-\frac{8}{-3}\right)p=-\frac{20}{-3}$$

Divide $-8$ by $-3$.

$$p^{2}+\frac{8}{3}p=-\frac{20}{-3}$$

Divide $-20$ by $-3$.

$$p^{2}+\frac{8}{3}p=\frac{20}{3}$$

Divide $\frac{8}{3}$, the coefficient of the $x$ term, by $2$ to get $\frac{4}{3}$. Then add the square of $\frac{4}{3}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$p^{2}+\frac{8}{3}p+\left(\frac{4}{3}\right)^{2}=\frac{20}{3}+\left(\frac{4}{3}\right)^{2}$$

Square $\frac{4}{3}$ by squaring both the numerator and the denominator of the fraction.

$$p^{2}+\frac{8}{3}p+\frac{16}{9}=\frac{20}{3}+\frac{16}{9}$$

Add $\frac{20}{3}$ to $\frac{16}{9}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$p^{2}+\frac{8}{3}p+\frac{16}{9}=\frac{76}{9}$$

Factor $p^{2}+\frac{8}{3}p+\frac{16}{9}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(p+\frac{4}{3}\right)^{2}=\frac{76}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p+\frac{4}{3}\right)^{2}}=\sqrt{\frac{76}{9}}$$

Simplify.

$$p+\frac{4}{3}=\frac{2\sqrt{19}}{3}$$ $$p+\frac{4}{3}=-\frac{2\sqrt{19}}{3}$$

Subtract $\frac{4}{3}$ from both sides of the equation.

$$p=\frac{2\sqrt{19}-4}{3}$$ $$p=\frac{-2\sqrt{19}-4}{3}$$