Let [ 0] be in R.

Question

Let $$\lambda  \ne 0$$  be in R. If $$\alpha $$ and $$\beta $$ are the roots of the
equation, x2 - x + 2$$\lambda $$ = 0 and $$\alpha $$ and $$\gamma $$ are the roots of
the equation, $$3{x^2} - 10x + 27\lambda  = 0$$, then $${{\beta \gamma } \over \lambda }$$  is equal to:

Accepted Answer

$$\alpha $$ and $$\beta $$ are the roots of the
equation x2 - x + 2$$\lambda $$ = 0 .....(1)

$$ \therefore $$ $$\alpha  + \beta  = 1,\,\alpha \beta  = 2\lambda $$

$$\alpha $$ and $$\gamma $$ are the roots of
the equation, $$3{x^2} - 10x + 27\lambda  = 0$$ ......(2)

$$ \therefore $$ $$\alpha  + \gamma $$ = $${{10} \over 3}$$, $$\alpha \gamma $$ = $${{27\lambda } \over 3}$$ = 9$$\lambda $$

Multiplying equation (1) by 3 and subtracting form equation (2) we get

-7x + 21$$\lambda $$ = 0

$$ \Rightarrow $$ x = 3$$\lambda $$

$$ \therefore $$ $$\alpha $$ = 3$$\lambda $$

As $$\alpha $$ is root of equation (1) so

$$\alpha $$2 - $$\alpha $$ +  2$$\lambda $$ = 0

$$ \Rightarrow $$ 9$$\lambda $$2 - 3$$\lambda $$ +  2$$\lambda $$ = 0

$$ \Rightarrow $$ $$\lambda $$ = $${1 \over 9}$$

$$ \Rightarrow $$ $$\alpha $$ = 3$$ \times $$ $${1 \over 9}$$ = $${1 \over 3}$$

Also $$\alpha \beta  = 2\lambda $$ = $${2 \over 9}$$

$$ \Rightarrow $$ $$\beta $$ = $${2 \over 3}$$

Also $$\alpha \gamma $$  = 9$$\lambda $$ = 9$$ \times $$ $${1 \over 9}$$ = 1

$$ \Rightarrow $$ $$\gamma $$ = 3

$$ \therefore $$ $${{\beta \gamma } \over \lambda }$$ = $${{{2 \over 3}.3} \over {{1 \over 9}}}$$ = 18

Solutions · 1