Question

How do you factor and solve $6{{x}^{2}}-9x=0$?

Answer 1

Hint: In the given question, we have been given a quadratic equation. We will first take the common terms out present in the expression and then equate the terms to zero. We will then get two values of \[x\]. Hence, we will have two real roots of the given quadratic equation.

Complete step by step solution:
According to the given question, we are provided with a quadratic equation which we have to solve in terms of \[x\].
Let’s start with writing the given expression, we have,
\[6{{x}^{2}}-9x=0\]----(1)
If we look carefully at the above equation, we can see that the equation has multiple of 3 in both the terms in the equation. So, we will take 3 out as a common factor. We get,
\[\Rightarrow 3(2{{x}^{2}}-3x)=0\]
Now, we can also see that from the obtained expression, we can take \[x\] common out, we get,
\[\Rightarrow 3x(2x-3)=0\]
Now, equating the terms to zero, we have,
\[3x=0\] or \[2x-3=0\]
We get,
\[x=0,\dfrac{3}{2}\]
Therefore, the values of \[x\] are \[0\] and \[\dfrac{3}{2}\].

Note:
If we want to check that the obtained values of \[x\] is correct or not. We have a way of doing that. We can simply substitute the obtained values in the given quadratic expression. If the expression gives us a value of 0, we have the correct answer else if we had made any mistakes while solving, the expression will give a value other than 0.
So, putting \[x=0\] in the equation (1), we get,
\[6{{x}^{2}}-9x=0\]
Putting the value in the LHS, we get,
\[\Rightarrow 6{{(0)}^{2}}-9(0)\]
\[\Rightarrow 0\]
Since, we got the value as 0, so the answer \[x=0\] is correct.
Next we have, \[x=\dfrac{3}{2}\],
Putting in equation (1), we get,
\[\Rightarrow 6{{\left( \dfrac{3}{2} \right)}^{2}}-9\left( \dfrac{3}{2} \right)\]
\[\Rightarrow 6\left( \dfrac{9}{4} \right)-9\left( \dfrac{3}{2} \right)\]
\[\Rightarrow 3\left( \dfrac{9}{2} \right)-9\left( \dfrac{3}{2} \right)=0\]
Therefore, the values of \[x\] which are \[0\] and \[\dfrac{3}{2}\] are correct.