Question

How do you solve $ 5{{x}^{2}}+25x=0 $ ?

Answer 1

Hint: We are given $ 5{{x}^{2}}+25x=0 $ , to solve this we will learn about quadratic equations, number of solutions of a quadratic equation. We will learn how to factorize the quadratic equation, we will simplify by taking the common terms out then we will use zero product rule to get our answer. At last, we will also learn about the quadratic formulas for solving such equations which is an easy and a more speedy way.

Complete step by step answer:
We are given $ 5{{x}^{2}}+25x=0 $ , we are asked to solve the given equation. First, we observe that it has a maximum power of 2, so it is a quadratic equation.
Now, we should know that a quadratic equation has a solution or we can say an equation of power n, will have n solution.
Now, we have to solve the equation $ 5{{x}^{2}}+25x=0 $ . To solve this equation, we will first take the greatest common factor possibly available to both the terms. As we can see that in $ 5{{x}^{2}}+25x=0 $ , the first term is $ 5{{x}^{2}} $ which can be written as $ 5\times x\times x $ and the other term is $ 25x $ which can be written as $ 5\times 5\times x $ .
Now, we can see that in both the terms, 5 and x are common.
 $ 5{{x}^{2}}+25x=5x\left( x+5 \right)=0 $
So, we get that the product of 5x and (x+5) gives us zero.
Now, we will use the zero product rule which says that if the product of a and b is zero, then either a or b will be zero.
Here, as 5x and (x+5) are the products, so either 5x = 0 or (x+5) = 0.
When we consider 5x = 0, we get x = 0.
And when we consider x+5 = 0, we get x = -5.
So, we get x = 0 and x = -5 as the two solution of the equation $ 5{{x}^{2}}+25x=0 $ .

Note:
 Another method to solve this problem is by using the quadratic formula which says that for an equation, $ 5{{x}^{2}}+25x=0 $ , the solution is given as $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $ . As we have our equation as $ 5{{x}^{2}}+25x=0 $ , here we will get a=5, b=25 and c=0. So, by using the quadratic formula, we will get,
 $ x=\dfrac{-25\pm \sqrt{{{25}^{2}}-4\times 5\times 0}}{2\times 5} $
On simplifying, we will get,
 $ x=\dfrac{-25\pm \sqrt{{{25}^{2}}}}{10} $
We know that \[\sqrt{{{25}^{2}}}=25\] so, we get,
 $ x=\dfrac{-25\pm 25}{10} $
So, our two solutions will be obtained after simplification. So, we get the solution as,
 $ \begin{align}
  & x=\dfrac{-25+25}{10},x=\dfrac{-25-25}{10} \\
 & \Rightarrow x=\dfrac{0}{10},x=\dfrac{-50}{10} \\
 & \Rightarrow x=0,x=-5 \\
\end{align} $
Hence, we get the solution of $ 5{{x}^{2}}+25x=0 $ as x = 0 and x = -5.