Question

How do you solve $5{{p}^{2}}-25=4{{p}^{2}}+24$?

Answer 1

Hint: We will take the help of a shifting method in order to solve this question. In this type of method, we have to keep variables and constants on either side of the equal sign. Since, there is a square variable in the given equation so, we will use the square root of 49 as 7 and solve it further. As the square root will give answers with both positive and negative signs so, we will consider them both as the right answer.

Complete step by step solution:
We will use a shifting process to solve this question. For this we are going to keep variables on one side and constants on the other. Now, we will consider the given equation $5{{p}^{2}}-25=4{{p}^{2}}+24$ and keep variables on left side and rest of constants on the right side of equal sign. Thus, we get
$\begin{align}
  & 5{{p}^{2}}-25=4{{p}^{2}}+24 \\
 & \Rightarrow 5{{p}^{2}}-4{{p}^{2}}=+24+25 \\
 & \Rightarrow 1{{p}^{2}}=49 \\
 & \Rightarrow {{p}^{2}}=49 \\
 & \Rightarrow p=\pm \sqrt{49} \\
\end{align}$
Since, the square root of 49 is 7 therefore, we get that p = + 7 or – 7.

Hence, the required value of p is either +7 or – 7.

Note: In order we need to be sure that our answer is correct or not we will cross check the given answer by substituting them in place of p. Since both 7 and – 7 will satisfy the given equation so, our answer is correct. There is no alternative method to solve this question. In case we try to do so, then we would first factorize the numbers and then take out the common number. This is shown below.
$\begin{align}
  & 5{{p}^{2}}-25=4{{p}^{2}}+24 \\
 & \Rightarrow 5\left( {{p}^{2}}-5 \right)=4\left( {{p}^{2}}+6 \right) \\
\end{align}$
After this we can cross multiply the terms together and get the following answer.
$\Rightarrow \dfrac{5}{4}=\dfrac{{{p}^{2}}-5}{{{p}^{2}}+6}$
Since, this leads to nowhere so, we have not applied it in this question.