Question

There is only one real value for ‘a’ for which the quadratic equation $a{{x}^{2}}+\left( a+3 \right)x+a-3=0$ has two positive integral solutions. The product of these two solutions is
(a) 9
(b) 8
(c) 6
(d) 12

Answer 1

To find the product of the two solutions of the given quadratic equation, we will first try to find the nature and value of ‘a’. For that, we will solve the quadratic equation with the help of the given information. The two key points mentioned in the question are (I) ‘a’ has only one real value and (II) the given equation has two positive integral solutions. Keeping these in focus, we proceed to solve the question.

Complete step-by-step solution:
We know that for a quadratic equation $a{{x}^{2}}+bx+c=0$ with roots ${{x}_{1}}$ and ${{x}_{2}}$,
Sum of roots $={{x}_{1}}+{{x}_{2}}=\dfrac{-b}{a}$
Product of roots $={{x}_{1}}{{x}_{2}}=\dfrac{c}{a}$
We know that the sum and product of two positive integers is always a positive integer. Then, the sum and product of the roots of the given equation are greater than zero and an integer.
Thus for the given equation $a{{x}^{2}}+\left( a+3 \right)x+a-3=0\text{ }\ldots (i)$
Sum of roots $=\dfrac{-\left( a+3 \right)}{a}>0\Rightarrow -1-\dfrac{3}{a}>0$
Product of roots $=\dfrac{a-3}{a}>0\Rightarrow 1-\dfrac{3}{a}>0$
The above two inequalities can only be satisfied when $a<0\text{ }\ldots (ii)$

So, $\dfrac{-\left( a+3 \right)}{a}\in \mathbb{Z}\Rightarrow -1-\dfrac{3}{a}\in \mathbb{Z}\Rightarrow \dfrac{3}{a}\in \mathbb{Z}$
And, $\dfrac{a-3}{a}\in \mathbb{Z}\Rightarrow 1-\dfrac{3}{a}\in \mathbb{Z}\Rightarrow \dfrac{3}{a}\in \mathbb{Z}$
Let us assume, $p=\dfrac{3}{a}\Rightarrow a=\dfrac{3}{p}\text{ }\ldots \left( iii \right)$
Putting this value in equation $(i)$, we get
$\dfrac{3}{p}{{x}^{2}}+\left( \dfrac{3}{p}+3 \right)x+\dfrac{3}{p}-3=0$
Solving, we get

& \dfrac{3}{p}{{x}^{2}}+\left( \dfrac{3+3p}{p} \right)x+\dfrac{3-3p}{p}=0 \\\ & \Rightarrow \dfrac{3}{p}{{x}^{2}}+\dfrac{3}{p}\left( 1+p \right)x+\dfrac{3\left( 1-p \right)}{p}=0 \\\ & \Rightarrow \dfrac{3}{p}\left[ {{x}^{2}}+\left( 1+p \right)x+\left( 1-p \right) \right]=0 \\\ & \Rightarrow {{x}^{2}}+\left( 1+p \right)x+\left( 1-p \right)=0\text{ }\ldots \left( iv \right) \\\ \end{aligned}$$ We know, for a quadratic equation $a{{x}^{2}}+bx+c=0$, the roots can be found with the help of the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ Then equation $\left( iv \right)$ can be solved as $$\begin{aligned} & x=\dfrac{-\left( 1+p \right)\pm \sqrt{{{\left( 1+p \right)}^{2}}-4\left( 1-p \right)}}{2} \\\ & =\dfrac{-\left( 1+p \right)\pm \sqrt{1+{{p}^{2}}+2p-4+4p}}{2} \\\ & =\dfrac{-\left( 1+p \right)\pm \sqrt{{{p}^{2}}+6p-3}}{2} \\\ & =\dfrac{-\left( 1+p \right)\pm \sqrt{{{\left( p+3 \right)}^{2}}-12}}{2} \end{aligned}$$ Since, the given equation $\left( i \right)$ has two positive integral solutions, Then, $$\sqrt{{{\left( p+3 \right)}^{2}}-12}$$ must be a perfect square. The perfect square closest to 12 is 16. Thus, $$\begin{aligned} & {{\left( p+3 \right)}^{2}}=16 \\\ & \Rightarrow p+3=\pm 4 \\\ & \Rightarrow p=4-3,-4-3 \\\ & \Rightarrow p=1,-7 \\\ \end{aligned}$$ Putting these values in equation $\left( iii \right)$, we get If $p=1\Rightarrow a=\dfrac{3}{1}=3$ And, $p=-7\Rightarrow a=\dfrac{3}{-7}=-\dfrac{3}{7}$ By using equation $\left( ii \right)$ , we get $a=-\dfrac{3}{7}$ Thus, product of roots of equation $\left( i \right)=\dfrac{a-3}{a}=\dfrac{\dfrac{-3}{7}-3}{\dfrac{-3}{7}}=\dfrac{-3-21}{-3}=\dfrac{-24}{-3}=8$ **Hence, option (b) is correct.** **Note:** While solving such sort of questions, some hit, and trial method is applied in rough. Like while solving for p in the above question, we concluded that $$\sqrt{{{\left( p+3 \right)}^{2}}-12}$$must be a perfect square, as otherwise if it is not a perfect square, it will lead to rational or irrational values of x which contradict the information given in the question. And, the value of $${{\left( p+3 \right)}^{2}}>12\Rightarrow {{\left( p+3 \right)}^{2}}$$ must be 16 as it is a square of a number.