Question

How do you solve $5{{p}^{2}}-15p=0$?

Answer 1

Hint: In the question, we are given the quadratic equation $5{{p}^{2}}-15p=0$ to solve. We can see that the factor of $5$ is common to the terms in the LHS, which means that it can be taken common and shifted to the RHS so that the given equation will get reduced to ${{p}^{2}}-3p=0$. Then, we can also take the factor of p common from the LHS so that it will get factored and we will get $p\left( p-3 \right)=0$. This equation can easily be solved for p using the zero product rule and equating each of the two factors to zero.

Complete step by step solution:
The quadratic equation in the above question is given as
$\Rightarrow 5{{p}^{2}}-15p=0$
We can observe in the LHS of the above equation that the factor of $5$ is common to both of the terms so that we can take it common in the LHS to get
\[\Rightarrow 5\left( {{p}^{2}}-3p \right)=0\]
Now, we can divide the above equation by \[5\] to get
\[\Rightarrow {{p}^{2}}-3p=0\]
We can now see that in the LHS of the above equation, the factor of p is also common. So we can take p common from the LHS to get
\[\Rightarrow p\left( p-3 \right)=0\]
Now, finally using the zero product rule, we can write
$\Rightarrow p=0$
And
$\begin{align}
  & \Rightarrow p-3=0 \\
 & \Rightarrow p=3 \\
\end{align}$
Hence, we have finally got the solutions of the given equation as $p=0$ and $p=3$.

Note: We can also use the quadratic formula to solve the given equation, which is given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. From the given equation $5{{p}^{2}}-15p=0$, we can note the coefficients as $a=5,b=-15,c=0$. On substituting these into the quadratic formula, we will get the solutions of the given equation.