Question

The roots of the equation
$9 {x}^{2} + 6x + 1 = 4kx$
where
$k$
is a real constant are denoted by
$\alpha$
and
$\beta$
a) show that the equation chose roots are

$\frac{1}{ \alpha }$
and
$\frac{1}{ \beta }$
is
$x {}^{2} + 6x + 9 = 4kx$
b) Find the est of values of
$k$
for which
$\alpha$
and
$\beta$

are real.
c) Find also the set of values of
$k$
for which
$\alpha$
and
$\beta$
are real and positive.
​

Answer 1

Answer: Correct option is

B

3x  

2

+12x+5

Let π:x  

3

−3x  

2

−4x+12=0

Sum of roots taken one at a time of π

⇒α+β+γ=3

of f(x)=(α−3)+(β−3)+(γ−3)=3−9=−6  

⇒  

a

−b

​

=−6⇒b=6{a=1}

Sum of roots taken two at a time of π  

αβ+βγ+γα=−4

of f(x)=(α−3)(β−3)+−−

=αβ−3(α+β)+9+−−  

=(αβ+βγ+γα)−3(α+β+β+γ+γ+α)+27

=−4−3(6)+27=5

⇒  

a

c

​

=5⇒c=5

⇒f(x) is of from x  

3

+6x  

2

+5x+d=0

f  

′

(x)=3x  

2

+12x+5⇒(B)

Step-by-step explanation: