Question

How do you solve $2{{x}^{2}}+9x+10=0$?

Answer 1

Hint: The given equation $2{{x}^{2}}+9x+10=0$ is a quadratic equation in the variable $x$. So we can use the middle term splitting technique to solve it. For this, we have to split the middle term $9x$ into the sum of two terms whose product must be equal to the product of the first and the third terms, that is, equal to $20{{x}^{2}}$. So we will split the middle term as \[9x=4x+5x\] and write the equation as \[2{{x}^{2}}+4x+5x+10=0\]. Then on taking the common factors outside from each of the pairs of the first two and the last two terms, we will be able to factorize the equation which can then be easily solved.

Complete step by step solution:
The given equation is
$\Rightarrow 2{{x}^{2}}+9x+10=0$
Using the middle term splitting technique, we have to split the middle term $9x$ as a sum of two terms whose product is equal to the product of the terms \[2{{x}^{2}}\] and $10$, that is equal to \[20{{x}^{2}}\]. So we split the middle term as \[9x=4x+5x\] to write the above equation as
$\Rightarrow 2{{x}^{2}}+4x+5x+10=0$
Now, taking $2x$ outside the first two terms, and $5$ outside the last two terms, we get
$\Rightarrow 2x\left( x+2 \right)+5\left( x+2 \right)=0$
Now, we take the factor $\left( x+2 \right)$ common to get
$\Rightarrow \left( x+2 \right)\left( 2x+5 \right)=0$
Using the zero product rule we get
$\begin{align}
  & \Rightarrow x+2=0 \\
 & \Rightarrow x=-2 \\
\end{align}$
And
$\begin{align}
  & 2x+5=0 \\
 & \Rightarrow x=-\dfrac{5}{2} \\
\end{align}$

Hence, the solutions of the given equation $x=-2$ are and $x=-\dfrac{5}{2}$

Note: We must note that the middle term $9x$ cannot be split in any arbitrary way. It has to be split into two terms such that their product is equal to the product of the first term $\left( 2{{x}^{2}} \right)$ and the last term $\left( 10 \right)$. If you are not able to guess a perfect split for the middle term, then you can also use the quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ in which substituting the values of the coefficients, you will directly get the solutions.