Question

please anwser it fastly.......

if a/k, a, ak are the roots of the equation x^3-px^2+qx-r=0 then a=​

Answer 1

✰Answer✰

♦️Given:

• a/k, a, ak are the zeroes of cubic polynomial x^3- px^2+qx-r=0.

♦️To find:

• Value of a....?

✰You must know✰

☛Let α, β, γ are the zeroes of cubic polynomial f(x) = ax^3 + bx^2 + cx + d = 0, Then

$\large{ \sf{ \star{sum \: of \: zeroes = - \frac{coefficient \: of \: {x}^{2} }{coefficient \: of \: {x}^{3} } }}} \\ \\ \large{ \sf{ \star{ sum \: of \: product \: of \: zeroes \: taken \: two \: at \: a \: time \: = \frac{coefficient \: of \: x}{coefficient \: of \: {x}^{3} } }}} \\ \\ \large{ \sf{ \star{product \: of \: zeroes = - \frac{constant \: term}{coefficient \: of \: of \: {x}^{3} } }}}$

☛With Reference to α,β,γ

$\large{ \sf{ \circ{ \: \: \: \alpha + \beta + \gamma = \frac{ - b}{a} }}} \\ \\ \large{ \sf{ \circ{ \: \: \: \alpha \beta + \beta \gamma + \alpha \gamma = \frac{ c}{a} }}} \\ \\ \large{ \sf{ \circ{\: \: \: \alpha \beta \gamma = \frac{ - d}{a} }}}$

✰Solution✰

♦️According to question

f(x) = x^3-px^2+qx-r

Zeroes of f(x) = a/k, a, ak.

☛By using the concept above,

$\large{ \sf{ \rightarrow \: \: product \: of \: zeroes = - \frac{constant \: term}{coefficient \: of \: {x}^{3} } }} \\ \\ \large{ \sf{ \star{ \: \: f(x) = \red{ \underline{{x}^{3}}} - p {x}^{2} + qx \red{ \underline{- r}}}}} \\ \\ \large{ \sf{ \rightarrow \: \frac{a}{ \cancel{k}} \times a \times a \cancel{k} = \frac{ - r}{1} }} \\ \\ \large{ \sf{ \rightarrow \: {a}^{3} = - r}} \\ \\ \large{ \sf{ \boxed{ \rightarrow \: \boxed{ \purple{a = \sqrt[3]{ - r} }}}}}$

✰So final Answer✰

$\large{ \boxed{ \bf{a = \sqrt[3]{ - r} }}}$