Copyright

Solve the equation. 1+\frac{1}{6p}=2~(p+5)

Question:

Solve the equation.

{eq}\displaystyle 1 + \frac{1}{6p} = 2~(p+5) {/eq}

Quadratic Equation:

An equation which is of the form {eq}ax^2+bx+c=0 {/eq} where {eq}a, b \text{ and } c {/eq} are constants is called a quadratic equation in {eq}x {/eq}. It has two (either same or distinct) roots. The roots can be solved by using the quadratic formula which states:

$$x=\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} $$

Answer and Explanation:

The given equation is:

$$1+\frac{1}{6p}=2~(p+5) $$

Multiply each term on both sides by {eq}6p {/eq}:

$$6p+1 = 2(p+5)(6p) \\[0.5cm] 6p+1= 12p(p+5)\\[0.5cm] 6p+1= 12p^2+60p \\[0.5cm] \text{Subtracting each of 6p and 1 from both sides},\\[0.5cm] 12p^2+54p-1=0 $$

Comparing this with {eq}ap^2+bp+c=0 {/eq}:

$$a=12\\ b=54\\ c=-1 $$

Substitute all these values in the quadratic formula:

$$\begin{align} p &=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\\[0.5cm] &= \frac{-54 \pm \sqrt{54^{2}-4(12)(-1)}}{2 \cdot 12} \\[0.5cm] &= \frac{-54 \pm \sqrt{2964}}{2 \cdot 12} \\[0.5cm] &= \frac{-54 \pm 2\sqrt{741}}{2 \cdot 12} \\[0.5cm] &= \frac{-27\pm \sqrt{741}}{12}\\[0.5cm] &= \color{blue}{\boxed{\mathbf{ \frac{-27+\sqrt{741}}{12}; \,\,\, \frac{-27-\sqrt{741}}{12}}}} \end{align} $$


Learn more about this topic:

Loading...
How to Use the Quadratic Formula to Solve a Quadratic Equation

from Math 101: College Algebra

Chapter 4 / Lesson 10
53K

Related to this Question

Explore our homework questions and answers library