Question

class 12...
no irrelevant stuff.. no spamming
no guessing
need answer + explanation.....
need help....​

Answer 1

$\huge\bf\red{\underline{\underline{ANSWER:-}}}$

For a Cubic equation -

Given :-

$\sf{kx^{3} - 26x^{2} + 52x - 24 = 0}$

$\sf{roots \: of \: above \: cubic \: eq. \: are \: in \: G.P. }$

To Find :-

$\sf{what's \: the \: value \: of \: k .}$

We know :-

Form of a Cubic equation -

$\sf{ax^{3} + bx^{2} + cx + d = 0}$

On Comparing the giving Cubic equation from

this standard form of cubic eq. -

$\sf{a = k}$

$\sf{b = -26}$

$\sf{c = 52}$

$\sf{d = -24}$

For a Cubic equation -

$\sf{\alpha + \beta + \gamma = -\dfrac{b}{a}}$

$\sf{\alpha \beta + \beta \gamma + \gamma \alpha =\dfrac{c}{a}}$

$\sf{\alpha \beta \gamma = -\dfrac{d}{a}}$

Solution :-

$\sf{Roots \: of \: the \: Cubic \: eq. \: (\alpha \: , \: \beta \: , \: \gamma) \: , \: are \: in \: G.P. -}$

So ,

$\sf{\alpha = a}$

$\sf{\beta = ar}$

$\sf{\gamma = ar^{2}}$

Now ,

$\sf{(a)(ar)(ar^{2}) = -\dfrac{-24}{k}}$

$\sf{(a^{3}r^{3}) =\dfrac{24}{k}}$

$\sf{ar =\sqrt[3]{\dfrac{24}{k}} ---(1)}$

then ,

$\sf{a + ar + ar^{2} =-\dfrac{-26}{k}}$

$\sf{a(1 + r + r^{2}) =\dfrac{26}{k}}$

$\sf{(1 + r + r^{2}) =\dfrac{26}{ka} --(2)}$

Now ,

$\sf{a(ar) + ar(ar^{2}) + ar^{2}(a) =\dfrac{52}{k}}$

$\sf{a^{2}r + a^{2}r^{3} + a^{2}r^{2} =\dfrac{52}{k}}$

$\sf{a^{2}r(1 + r + r^{2}) =\dfrac{26}{k}\times 2}$

From eq.(2) -

$\sf{ar[a(1 + r + r^{2})] =\dfrac{26}{k}\times 2}$

$\sf{(ar)(\dfrac{26}{k}) =\dfrac{26}{k}\times 2}$

$\sf{ar = 2}$

$\sf{\sqrt[3]{\dfrac{24}{k}}= 2}$

Doing Cube both side -

$\sf{\dfrac{24}{k} = 2^{3}}$

$\sf{\dfrac{24}{k} = 8}$

$\sf{k =\dfrac{24}{8}}$

$\sf{k = 3}$