Question

If $2a+3b+6c=0,$ then at least one root of the equation $a{x}^2+bx+c=0$ lies in the interval

• A. (0, 1)
• B. (1, 2)
• C. (2, 3)
• D. (1, 3)

Answer 1

Answer: (A)

Let $f(x)=a{x}^2+bx+c$ $\Rightarrow$ $f(x)=\frac{a{x}^3}{3}+\frac{b{x}^2}{2}+cx+d$ $\Rightarrow$ $f(x)=\frac{2a{x}^3+3b{x}^2+6cx+6d}{6}$ $\Rightarrow$ $f(1)=\frac{2a+3b+6c+6d}{6}=\frac{6d}{6}=d$ $(\because 2a+3b+6c=0)$ $f(0)=\frac{6d}{6}=d$ $\therefore$ $f(0)=f(1)$ $\Rightarrow$ $f(x)$ will vanish at least once between 0 and 1. $\therefore$ One of the roots of $a{x}^2+bx+c=0$ lies between 0 and 1.