Question

How do you solve $10{{x}^{2}}+19x+6=0$ by factoring?

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 coefficient of and constant term and give the result of their multiplication equal to coefficient of x, that is, 19.

Complete step-by-step answer:
Since, the method of factoring the quadratic equation makes our calculations simpler, therefore, we use it the most.
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, $10{{x}^{2}}+19x+6=0$, $a=10,$ $b=19$ and $c=6$.
We will find numbers by hit and trial whose product is equal to $10\times 6=60$ and whose sum is equal to $19$.
Such two numbers are $15$ and $4$ as $15+4=19$ and $15\times 4=60$.
Now, factoring the equation:
$\Rightarrow 10{{x}^{2}}+15x+4x+6=0$
Taking common, we get:
$\begin{align}
  & \Rightarrow 5x\left( 2x+3 \right)+2\left( 2x+3 \right)=0 \\
 & \Rightarrow \left( 2x+3 \right)\left( 5x+2 \right)=0 \\
\end{align}$
Hence, $2x+3=0$ or $5x+2=0$
On transposing the constant terms to the other side and then dividing it with the coefficient of x, we get
$\Rightarrow x=-\dfrac{3}{2}$ or $x=-\dfrac{2}{5}$
Therefore, the roots of the given quadratic equation are $x=-\dfrac{3}{2},-\dfrac{2}{5}$.

Note:
Another method of solving this problem and finding the roots of the given quadratic equation is by first calculating the value of discriminant, D using the coefficients of the terms in the equation. Then, we shall find the both the roots of the equation. This method is usually used when the roots of the equation are irrational or complex numbers.