Question

Let S be the set of all non-zero real numbers $\alpha$such that the quadratic equation $\alpha {{x}^{2}}-x+a=0$ has two distinct real roots x1 and x2 satisfying the inequality $|{{x}_{1}}-{{x}_{2}}|<1$. Which of the following intervals is (are) subset(s) of S?
(a) $\left( \dfrac{-1}{2},\dfrac{-1}{\sqrt{5}} \right)$
(b) $\left( \dfrac{-1}{\sqrt{5}},0 \right)$
(c) $\left( 0,1 \right)$
(d) $\left( \dfrac{1}{\sqrt{5}},\dfrac{1}{2} \right)$

Answer 1

Hint: In this question, we first need to make the discriminant greater than 0. Then from the given inequality by substituting the relation between the roots we get another condition. Now, by considering both the conditions obtained above we get the subset.

Complete step-by-step answer:

Let us now consider the quadratic equation given in the question
$\alpha {{x}^{2}}-x+a=0$
Given that it has distinct real roots which means that the discriminant is greater than 0.
As we already know that the discriminant of a quadratic equation is given by the formula.
${{b}^{2}}-4ac$
Now, from the given quadratic equation in the question we have,

& b=-1 \\\ & a=\alpha \\\ & c=a \\\ \end{aligned}$$ Now, by substituting these values in the above discriminant formula and making it greater 0 according to the condition we get, $$\Rightarrow {{\left( -1 \right)}^{2}}-4\times \alpha \times a>0$$ Now, on further simplification we get, $$\Rightarrow 1-4\alpha a>0$$ As we already know that in a quadratic equation $$a{{x}^{2}}+bx+c=0$$ The sum of the roots is given by $$\dfrac{-b}{a}$$ The product of the roots is given by $$\dfrac{c}{a}$$ Now, by using these formulae from the given quadratic equation in the question we get, $$\begin{aligned} & {{x}_{1}}+{{x}_{2}}=\dfrac{1}{\alpha } \\\ & {{x}_{1}}\cdot {{x}_{2}}=\dfrac{a}{\alpha } \\\ \end{aligned}$$ Now, from the given inequality in the question we get, $$\Rightarrow |{{x}_{1}}-{{x}_{2}}|<1$$ Now, by squaring on both sides we get, $$\Rightarrow {{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}<1$$ Let us further rewrite it in terms of the known form. $$\Rightarrow {{\left( {{x}_{1}}+{{x}_{2}} \right)}^{2}}-4{{x}_{1}}\cdot {{x}_{2}}<1$$ Now, by substituting the respective values in the above inequality we get, $$\Rightarrow {{\left( \dfrac{1}{\alpha } \right)}^{2}}-\dfrac{4a}{\alpha }<1$$ Let us now further simplify this $$\Rightarrow \dfrac{1-4\alpha a}{{{\alpha }^{2}}}<1$$ Now, from the condition obtained from the discriminant the numerator will be greater than 0 and the denominator will be always positive. So, we can further write it as $$\Rightarrow 0<\dfrac{1-4\alpha a}{{{\alpha }^{2}}}<1$$ Therefore, $$S=\left( 0,1 \right)$$ Hence, the correct options are (c) and (d) which are subsets of S. Note: It is important to note that as the given quadratic equation has distinct roots which means that its discriminant is greater than 0. If we do not consider this condition we cannot get the subset because the given inequality gives only the maximum value possible but not the minimum value. It is also to be noted that while squaring the given inequality the inequality remains the same. If changed then the condition changes completely and so the subset cannot be found.