Question

Let $a,b(b>a)$ are the roots of the quadratic equation $(k+1)x^2-(20k+14)x+91k+40=0;$ where $k>0$, then which among the following option(s) is/are correct for the roots

• A. $a\in (4,7)$
• B. $b\in (4,7)$
• C. $a\in (7,10)$
• D. $b\in (10,13)$

Answer 1

Answer: (A), (D)

The correct options are
A $a\in (4,7)$
D $b\in (10,13)$

$(k+1)x^2-(20k+14)x+91k+40=0;k>0$
The equation can be rewritten as
$\Rightarrow x^2-\frac{20k+14}{k+1}x+\frac{91k+40}{k+1}=0$
Let $f\left(x\right)=x^2-\frac{20k+14}{k+1}x+\frac{91k+40}{k+1}$
$\Rightarrow f\left(x\right)=x^2-\left(\frac{20(k+1)-6}{k+1}\right)x+\frac{91(k+1)-51}{k+1}\Rightarrow f\left(x\right)=(x^2-20x+91)+\left(\frac{6x-51}{k+1}\right)\Rightarrow f\left(x\right)=(x-13)(x-7)+\left(\frac{6x-51}{k+1}\right)$

Now,
$f\left(4\right)=(-9)(-3)-\frac{27}{k+1}\Rightarrow f\left(4\right)=\frac{27k}{k+1}>0(\because k>0)f\left(7\right)=\frac{-9}{k+1}<0f\left(10\right)=(-3)\left(3\right)+\frac{9}{k+1}\Rightarrow f\left(10\right)\frac{-9k}{k+1}<0f\left(13\right)=\frac{27}{k+1}>0$
So, the graph of the polynomial


The equation has two roots- one in $(4,7)$ and another in $(10,13)$

Hence, as $b>a$ so $a\in (4,7),b\in (10,13)$