Question

The number of integers $n$ for which $3x^2-25x+n=0$ has three real roots is

• A. 1
• B. 25
• C. 55
• D. infinite

Answer 1

Answer: (C)

We have, $3x^2-25x+n=0$
Let $f\left(x\right)=3x^3-25x+n$
$f^{\prime }\left(x\right)=9x^2-25$
Put $f^{\prime }\left(x\right)=0$
$9x^2-25=0$
$x=\pm \frac{5}{3}$
$\Rightarrow x_1=\frac{-5}{3},x_2=\frac{5}{3}$
$f(x)$ has three real roots.
$\therefore f(x_1)f(x_2)<0$
$\therefore \left(3{\left(\frac{-5}{3}\right)}^3-25\left(\frac{-5}{3}\right)+n\right)$
$\left(3{\left(\frac{5}{3}\right)}^3-25\left(\frac{5}{3}\right)+n\right)<0$
$\left(\frac{-125}{9}+\frac{125}{3}+n\right)\left(\frac{125}{9}-\frac{125}{3}+n\right)<0$
$\left(n+\frac{250}{9}\right)\left(n-\frac{250}{9}\right)<0$
$n\in \left(\frac{-250}{9},\frac{250}{9}\right)$
Hence, $n\in I$
$\therefore$ Total integer is $55$.