Question

Find all the values of the parameter $d$ for which both roots of the equation $x^2-6dx+(2-2d+9d^2)=0$ exceed the number $3.$

• A. $(-\infty ,1)$
• B. $(\frac{11}{9},\infty )$
• C. $(-\infty ,1)$
• D. $(1,\frac{11}{9})$

Answer 1

The correct option is B $(\frac{11}{9},\infty )$

Let $f\left(x\right)=x^2-6dx+(2-2d+9d^2)$
If both roots exceed 3 then,
Conditions-
(i) $D\ge 0$

$\Rightarrow 36d^2-\mathrm{4.1.}(2-2d+9d^2)\ge 0$

$\Rightarrow 36d^2-8+8d-36d^2\ge 0$

$\Rightarrow 8d\ge 8\Rightarrow d\ge 1$ ........... (1)

(ii) $f\left(3\right)>0$

$\Rightarrow (3{)}^2-6d.3+2-2d+9d^2>0$

$\Rightarrow 9-18d+2-2d+9d^2>0$

$\Rightarrow (9d-11)(d-1)>0$

$d\in (-\infty ,1)\cup (\frac{11}{9},\infty )$ .......... (2)

(iii) $-\frac{b}{2a}>3\Rightarrow -\left(\frac{-6d}{2}\right)>3$

$\Rightarrow 3d>3\Rightarrow d>1$ ............ (3)

Taking intersection of (1), (2) and (3), we get

$d\in (\frac{11}{9},\infty )$