JEE Main 2020 (Online) 2nd September Evening ShiftQuestion 1
Let $f(x)$ be a quadratic polynomial such that $f(-1)+f(2)=0$. If one of the roots of $f(x)=0$ is $3$, then its other root lies in :
Quadratic Equation · Relation Between Roots and Coefficients
Solution
Concept. Write $f(x)=a(x-3)(x-r)$ and use the condition. Formulas used. (1) A quadratic with roots $3$ and $r$: $f(x)=a(x-3)(x-r)$, $a\ne0$Step 1 - Values. $f(-1)=4a(1+r)$ and $f(2)=-a(2-r)$. Step 2 - Condition. $a(4+4r-2+r)=a(5r+2)=0$, so $r=-\frac25$. Step 3 - Interval. $-\frac25\in(-1,0)$. Answer: (B)