Question

How do you solve $ {{x}^{2}}-36x=0 $ ?

Answer 1

Hint: We first try to take common terms out from the given equation $ {{x}^{2}}-36x=0 $ . We need to form factorisation from the left side equation $ {{x}^{2}}-36x $ . We have only the variable $ x $ to take as common. From the multiplication we find the solution for $ {{x}^{2}}-36x=0 $ .

Complete step-by-step answer:
We need to find the solution of the given equation $ {{x}^{2}}-36x=0 $ .
First, we try to take a common number or variable out of the terms $ {{x}^{2}} $ and $ -36x $ .
The only thing that can be taken out is $ x $ .
So, $ {{x}^{2}}-36x=x\left( x-36 \right)=0 $ .
The multiplication of two terms gives 0. This gives that at least one of the terms has to be zero.
We get the values of x as either $ x=0 $ or $ \left( x-36 \right)=0 $ .
This gives $ x=0,36 $ .
The given quadratic equation has 2 solutions and they are $ x=0,36 $ .
So, the correct answer is “ $ x=0,36 $ ”.

Note: The highest power of the variable or the degree of a polynomial decides the number of roots or the solution of that polynomial. Quadratic equations have 2 roots. Cubic polynomials have 3. It can be both real and imaginary roots.
We can also apply the quadratic equation formula to solve the equation $ {{x}^{2}}-36x=0 $ .
We know for a general equation of quadratic $ a{{x}^{2}}+bx+c=0 $ , the value of the roots of x will be $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $ .
In the given equation we have $ {{x}^{2}}-36x=0 $ . The values of a, b, c is $ 1,-36,0 $ respectively.
We put the values and get x as
$
 x=\dfrac{-\left( -36 \right)\pm \sqrt{{{\left( -36 \right)}^{2}}-4\times 1\times 0}}{2\times 1}
=\dfrac{36\pm \sqrt{{{36}^{2}}}}{2}
=\dfrac{36\pm 36}{2}=0,36
$ .