Question

How do you solve $2{{x}^{2}}-9x-35=0$?

Answer 1

Hint: For solving the equation $2{{x}^{2}}-9x-35=0$, we can use the middle term splitting method. For this, we first need to determine the product of the first term, $2{{x}^{2}}$ and the third term, $-35$. The product which we will obtain is equal to $-70{{x}^{2}}$. Then we need to split the middle term $-9x$ into two terms such that their product is equal to $-70{{x}^{2}}$. The only way to do this is to split it as $-9x=-14x+5x$. Then we will get a total four terms on the LHS of the equation from which two pairs of terms will be generated. On taking factors common from each pair, we will be able to factorize the LHS. The factored equation can be easily solved using the zero product rule.

Complete step by step solution:
The equation given in the above question is
$\Rightarrow 2{{x}^{2}}-9x-35=0$
Here, the product of the first and the third terms is $\left( 2{{x}^{2}} \right)\left( -35 \right)=-70{{x}^{2}}$. Therefore, by the middle term splitting method we split the middle term as $-9x=-14x+5x$ in the above equation to get
$\Rightarrow 2{{x}^{2}}-14x+5x-35=0$
So we have four terms. Let us make two pairs, one from the first two and the other from the last two terms.
$\Rightarrow \left( 2{{x}^{2}}-14x \right)+\left( 5x-35 \right)=0$
Now, we take $2x$ common from the first pair to get
$\Rightarrow 2x\left( x-7 \right)+\left( 5x-35 \right)=0$
Similarly, we take $5$ common from the second pair to get
$\Rightarrow 2x\left( x-7 \right)+5\left( x-7 \right)=0$
Now, the factor $\left( x-7 \right)$ is common. Taking it outside, we get
$\Rightarrow \left( x-7 \right)\left( 2x+5 \right)=0$
Using the zero product rule, we get
$\begin{align}
  & \Rightarrow x-7=0 \\
 & \Rightarrow x=7 \\
\end{align}$
And
$\begin{align}
  & \Rightarrow 2x+5=0 \\
 & \Rightarrow x=-\dfrac{5}{2} \\
\end{align}$

Hence, the solutions of the given equation are $x=7$ and $x=-\dfrac{5}{2}$.

Note: We can also solve the given equation by using the quadratic formula which is given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. From the given equation, the coefficients are $a=2$, $b=-9$ and $c=-35$. On substituting these values into the quadratic formula, we will get the same solutions as obtained above.