Question

How do you solve the equation for ${x^2} - 3x = 0$?

Answer 1

Hint: The given equation in the question is a quadratic equation. It is an equation containing a single variable of degree $2$. Its general form is $a{x^2} + bx + c = 0$, where $x$ is the variable and $a$, $b$, $c$ are the constants. Also, degree is the highest power (in whole number) that a variable has, in an equation.

Complete step by step solution:
Given quadratic equation is ${x^2} - 3x = 0$
Since it is a quadratic equation, therefore there will be two zeroes for the equation. It means that there will be two values of $x$ for which the equation ${x^2} - 3x = 0$ will be equals to zero. So, let us find the zeros of the equation by the method of factorization. We have,
$ \Rightarrow {x^2} - 3x = 0$
Here $x$ is a common factor of both the expressions ${x^2}$ and $ - 3x$, therefore we will separate it from them
$
   \Rightarrow x \times x - 3 \times x = 0 \\
   \Rightarrow x(x - 3) = 0 \\
 $
As we can see here, the product of $x$ and $x - 3$ is $0$. We know that the product of two numbers is zero when either one of them is zero. So here, we have
$ \Rightarrow x = 0$ or $ \Rightarrow x - 3 = 0$
$ \Rightarrow x = 0$ or $ \Rightarrow x = 3$

Hence, the solution for the equation ${x^2} - 3x = 0$ is $x = 0$ or $x = 3$.

Note: We have an alternate method to solve the given equation, i.e. with the help of quadratic formula.
The equation ${x^2} - 3x = 0$ is quadratic in nature and is hence in the form of $a{x^2} + bx + c = 0$. The quadratic formula is given as
${x_1} = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}$ and ${x_2} = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$ where ${x_1}$ and ${x_2}$ are the zeroes.
Here $a = 1$; $b = - 3$ and $c = 0$. On substituting these values in the formula, we get
$
  {x_1} = \dfrac{{ - ( - 3) + \sqrt {{{( - 3)}^2} - 4 \times 1 \times 0} }}{{2 \times 1}} \\
   \Rightarrow {x_1} = \dfrac{{3 + \sqrt 9 }}{2} \\
   \Rightarrow {x_1} = \dfrac{{3 + 3}}{2} = \dfrac{6}{2} \\
   \Rightarrow {x_1} = 3 \\
 $
And,
$
  {x_2} = \dfrac{{ - ( - 3) - \sqrt {{{( - 3)}^2} - 4 \times 1 \times 0} }}{{2 \times 1}} \\
   \Rightarrow {x_2} = \dfrac{{3 - \sqrt 9 }}{2} \\
   \Rightarrow {x_2} = \dfrac{{3 - 3}}{2} \\
   \Rightarrow {x_2} = 0 \\
 $
Hence, the solution for the equation ${x^2} - 3x = 0$ is ${x_1} = 3$ and ${x_2} = 0$.