Question

How do you solve $3{{x}^{2}}-x=10$?

Answer 1

Hint: We are given a quadratic equation which can be solved by the method of factoring the equation. We shall first find the sum of the roots and the product of the roots of the equation. Further we will find numbers which will add up to the sum of the roots and give the result of their multiplication as the product of their roots.

Complete step by step answer:
There are four methods for solving quadratic equations, namely, factoring method, completing the square method, taking the square root method and the last method is solving using the various properties of polynomials.
However, we will use the method of factoring the quadratic equation which makes our calculations simpler.
For any quadratic equation $a{{x}^{2}}+bx+c=0$,
the sum of the roots $=-\dfrac{b}{a}$ and the product of the roots $=\dfrac{c}{a}$.
Thus, for the equation, $3{{x}^{2}}-x=10$,
$\Rightarrow 3{{x}^{2}}-x-10=0$,
We will find numbers by hit and trial whose product is equal to $3\times \left( -10 \right)=-30$ and whose sum is equal to $-1$.
Such two numbers are $-6$ and $5$ as $-6+5=-1$ and $\left( -6 \right)\times 5=-30$.
Now, factoring the equation:
$\Rightarrow 3{{x}^{2}}-6x+5x-10=0$
Taking common, we get:
$\begin{align}
  & \Rightarrow 3x\left( x-2 \right)+5\left( x-2 \right)=0 \\
 & \Rightarrow \left( x-2 \right)\left( 3x+5 \right)=0 \\
\end{align}$
Hence, $x-2=0$ or $3x+5=0$
$\Rightarrow x=2$ or $x=-\dfrac{5}{3}$

Therefore, the roots of the equation are $x=2,-\dfrac{5}{3}$.

Note: We shall also get much more complex quadratic or bi-quadratic or even higher degree polynomials to be factorized. In that case, one must be careful enough to find the right common factors of the multiple terms given. We must also be careful enough while grouping the terms in order to avoid mistakes.