13 votes 13 votes The set of values of $p$ for which the roots of the equation $3x^2+2x+p(p–1) = 0$ are of opposite sign is $(–∞, 0)$ $(0, 1)$ $(1, ∞)$ $(0, ∞)$ Quantitative Aptitude gate2013-ee quantitative-aptitude quadratic-equations + – Akash Kanase asked Feb 16, 2016 • edited Jun 7, 2018 by Milicevic3306 Akash Kanase 3.3k views answer comment Share Follow See 1 comment See all 1 1 comment reply Desert_Warrior commented May 1, 2016 reply Follow Share Hi could u plz add this Q to numerical ability? 1 votes 1 votes Please log in or register to add a comment.
Best answer 24 votes 24 votes Roots of equation are of opposite sign then for the equation, $ax^2+bx+c =0$, product of roots, $\dfrac{c}{a}$, should be (negative number) less than $0$ $\dfrac{p(p-1)}{3}< 0$ $p(p-1)<0$ so $p$ must be less than $1$ and greater than $0$ Option B Praveen Saini answered Feb 16, 2016 • edited Dec 7, 2017 by pavan singh Praveen Saini comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments अनुराग पाण्डेय commented Feb 29, 2016 reply Follow Share @richa07 Consider a quadratic equation $ax^{2} + bx +c =0$, where $a, b$ and $c$ are constants and $x$ is variable. Roots of the quadratic equations are the value(s) of $x$ at which the equation $ax^{2} + bx +c =0$ holds or the value of the expression $ax^{2} + bx +c$ becomes zero. Without actually finding the roots of a quadratic equation we can infer some points about the nature of roots by just glancing over the quadratic equation. For example, if we have quadratic equation $ax^{2} + bx +c =0$, then without finding its roots, we can find the sum and product of roots, sum of roots would be $\frac{-b}{a}$ and product of roots would be $\frac{c}{a}$ It can also be verified by actually finding the roots(if possible) and then adding/multiplying them. Now if the two roots of a quadratic equation are of opposite signs then their product must be negative as the product of a negative number & a positive number would be a negative number. So we can say that the product of roots (i.e. $\frac{c}{a}$), which is given by $\frac{p\left ( p-1 \right )}{1}$ as $c = p\left ( p-1 \right )$ and $a = 1$ in the given quadratic equation, should be strictly less than zero. hence we need to solve $\frac{p\left ( p-1 \right )}{1}<0$, to find the interval(s) where roots of the given quadratic equation would have opposite sign. Following is the $p$ vs $p\left ( p-1 \right )$ plot, to show how the sign of the product of roots of the given quadratic equation are changing as we are varying the value of $p$ along the real number line. 24 votes 24 votes Praveen Saini commented Feb 29, 2016 reply Follow Share @Anurag: Very well explained 4 votes 4 votes अनुराग पाण्डेय commented Feb 29, 2016 reply Follow Share Thank you sir, my pleasure. :) 1 votes 1 votes Please log in or register to add a comment.
0 votes 0 votes Let the roots are C and -D. So we get C-D=2/3 and C*D=P(P-1)/3 should be <0 as the product of Cand D is negative as they are opposite signs. P(P-1)/3<0 p(P-1)<0 WHICH INDICATE S THE VALUE SHOULD LIE between 0 and 1 and the option B is correct. DIBAKAR MAJEE answered May 7, 2020 DIBAKAR MAJEE comment Share Follow See all 0 reply Please log in or register to add a comment.