Question

How do you solve the quadratic equation $23p=5{{p}^{2}}+24$?

Answer 1

Hint: The equation $23p=5{{p}^{2}}+24$ given in the above question is a quadratic equation in $p$. For solving it, we first have to convert it into the standard form of $a{{x}^{2}}+bx+c=0$ by subtracting the LHS $23p$ from both the sides to get the equation $5{{p}^{2}}-23p+24=0$. Then using the middle term splitting method we need to split the middle term $-23p$ into the two terms such that their product is equal to the product of the first term $5{{p}^{2}}$ and the third term $24$, which is equal to $120{{p}^{2}}$. Then taking the common terms outside from each pair of the four terms, we will be able to factorize the equation and the final solution will be obtained by using the zero product rule.

Complete step by step solution:
The given equation is
$\Rightarrow 23p=5{{p}^{2}}+24$
The given equation is a quadratic equation, which is not written in the standard form. Therefore, we subtract $23p$ from both the sides to get
\[\begin{align}
  & \Rightarrow 23p-23p=5{{p}^{2}}+24-23p \\
 & \Rightarrow 0=5{{p}^{2}}+24-23p \\
 & \Rightarrow 5{{p}^{2}}-23p+24=0 \\
\end{align}\]
Now, the product of the first and the third term is equal to $\left( 5{{p}^{2}} \right)\left( 24 \right)=120{{p}^{2}}$. Using the middle term splitting method, we split the middle term $-23p$ as $-23p=-15p-8p$ in the above equation to get
$\Rightarrow 5{{p}^{2}}-15p-8p+24=0$
Taking $5p$ common from the first two terms, and $-8$ common from the last two terms, we get
$\Rightarrow 5p\left( p-3 \right)-8\left( p-3 \right)=0$
Now, taking $\left( p-3 \right)$ common we get
$\Rightarrow \left( p-3 \right)\left( 5p-8 \right)=0$
Using the zero product rule, we get
$\begin{align}
  & \Rightarrow p-3=0 \\
 & \Rightarrow p=3 \\
\end{align}$
And
$\Rightarrow 5p-8=0$
Adding $8$ both sides
$\begin{align}
  & \Rightarrow 5p-8+8=0+8 \\
 & \Rightarrow 5p=8 \\
\end{align}$
Dividing by $5$ both sides, we finally get
$\begin{align}
  & \Rightarrow \dfrac{5p}{5}=\dfrac{8}{5} \\
 & \Rightarrow p=\dfrac{8}{5} \\
\end{align}$
Hence, the solutions of the given equation are $p=3$ and $p=\dfrac{8}{5}$.

Note: For solving any quadratic equation, it must be written in the standard form, that is, the RHS must be equal to zero. After writing the given equation into the standard form as $5{{p}^{2}}-23p+24=0$, we can also use the quadratic formula to solve it. The quadratic formula given by $p=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ is helpful when we are not able to split the middle term properly.