Question

If the equation $\left( a-5 \right){{x}^{2}}+2\left( a-10 \right)x+a+10=0$ has roots of opposite sign, then,
A. $a> 10$
B. $-15< a <15$
C. $-10< a< 5$
D. None of these.

Answer 1

Hint: Assume one root to be $\alpha $, therefore the other root will be $-\alpha $. Now we will apply the formula of sum of roots of a quadratic equation $\left[ \dfrac{-b}{a} \right]$ to solve further and get the desired answer.

Complete step-by-step solution:
Let us assume one of the roots to be $\alpha $, therefore we will get the other root as $-\alpha $as we are given that the equation has roots of opposite sign. Now we also know that,
Sum of roots of a quadratic equation = $\left[ \dfrac{-b}{a} \right]$, where - b is the coefficient of x and a is the coefficient of ${{x}^{2}}$. Therefore we can say that in the given equation, $\left( a-5 \right){{x}^{2}}+2\left( a-10 \right)x+a+10=0$, the coefficient of $x=2\left( a-10 \right)$ and the coefficient of ${{x}^{2}}=\left( a-5 \right)$. Now, we know that the,
$\text{sum of roots = }\dfrac{\text{- coefficient of x}}{\text{coefficient of }{{\text{x}}^{\text{2}}}}$
So, by substituting the values of coefficient of x and coefficient of ${{x}^{2}}$, we will get,
$\begin{align}
  & \alpha +\left( -\alpha \right)=\dfrac{-2\left( a-10 \right)}{\left( a-5 \right)} \\
 & \Rightarrow \alpha -\alpha =\dfrac{-2\left( a-10 \right)}{\left( a-5 \right)} \\
 & \Rightarrow 0=\dfrac{-2a+20}{a-5} \\
\end{align}$
We will now transfer $\left( a-5 \right)$ to the left hand side or the LHS using cross multiplication. So, we get,
\[\begin{align}
  & -2a+20=0 \\
 & \Rightarrow 2a=20 \\
 & \Rightarrow a=\dfrac{20}{2} \\
 & \Rightarrow a=10 \\
\end{align}\]
Therefore, when we check the options that are provide to us, we find that the most suitable one is $-15< a < 15$.
Hence, the correct answer is option B.

Note: The students should not make mistakes by getting confused with the formula of sum of roots and product of roots, which is $\dfrac{c}{a}$, where c is the constant and a is the coefficient of the term ${{x}^{2}}$ in the quadratic equation. The students should also watch their calculations as according to the standard equation, $a{{x}^{2}}+bx+c$ and in the given equation, we have a as the common variable, so don’t get confused by that.