Question

How do you solve ${{x}^{2}}-14x=0$ ?
 (a) Using Sridharacharya’s formula
(b) Simplifying the problem
(c) Separating the equations
(d) All of the above

Answer 1

Hint: According to the question, we are to find the value of x from the given equation ${{x}^{2}}-14x=0$. And we will start with analyzing the options one by one and choose which one of them is the right one. We will use the formula of Sridharacharya for an equation $a{{x}^{2}}+bx+c=0$ we have a solution as, $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . With that we can simplify the equation to get our right option.

Complete step by step answer:
To start with, we have the equation, ${{x}^{2}}-14x=0$
In the standard form of Sridharacharya’s formula we get, a = 1, b = -14 and c = 0,
So, for an equation $a{{x}^{2}}+bx+c=0$ we have a solution as, $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ ,
Now, putting the values, we are getting, $x=\dfrac{-\left( -14 \right)\pm \sqrt{{{\left( -14 \right)}^{2}}-4.1.0}}{2.1}$
Trying to simplify, $x=\dfrac{14\pm \sqrt{196}}{2}$
Again, $x=\dfrac{14\pm 14}{2}$
So, we get the value of x as, $\dfrac{28}{2}=14$ and $\dfrac{0}{2}=0$
Now, to get with the options, again, we have, the left hand side as, ${{x}^{2}}-14x=x\left( x-14 \right)$Thus, it can be written that, $\Rightarrow x\left( x-14 \right)=0$
This gives us x = 0 and x = 14.
By simplifying we also get the value of x as, $x=\dfrac{14+14}{2}$ and $x=\dfrac{14-14}{2}$
Thus, we also get our solutions of the equation as , x = 14 and x = 0.
Hence, we can say that ${{x}^{2}}-14x=0$ can be simplified to give the solutions as x = 14 and x = 0 by using Sridharacharya’s formula.

So, the correct answer is “Option a”.

Note: This problem is a comparatively easier one, which is going to be a problem to get the right solution. Easier problems are prone to making silly mistakes. So, we need to take care of the problem so that we do not make any calculation mistakes.