Question

if A and B are roots of 8x²+3x+2=0 then find the,value of (A²by B)⅓+(B²bA)⅓​

Answer 1

Question :–

▪︎ if A and B are roots of 8x² + 3x + 2 = 0 then find the value of $\: \: { \bold{ { ( \dfrac{ {A}^{2} }{B}) }^{ \frac{1}{3} } + {( \dfrac{ {B}^{2} }{A}) }^{ \frac{1}{3} } = ? }}$

ANSWER :–

$\\ \: \longrightarrow { \red { \boxed{ \bold{ { ( \dfrac{ {A}^{2} }{B}) }^{ \frac{1}{3} } + {( \dfrac{ {B}^{2} }{A}) }^{ \frac{1}{3} } = \frac{ - 3}{ {2}^{ \frac{7}{3} } } }}}}$

EXPLANATION :–

• A quadratic equation ax² + bx + c = 0 which have two roots A and B.

• So that –

$\\ \: \: \: { \huge{.}}{ \bold{ \: \: Sum \: \: of \: \: roots = A + B = - \frac{b}{a} }}$

$\\ \: \: \: { \huge{.}}{ \bold{ \: \: Product \: \: of \: \: roots = A . B = \frac{c}{a} }}$

• In given equation –

$\\ \: \: \: { \huge{.}}{ \blue{ \bold{ \: \: \: a = 8 }}}$

$\\ \: \: \: { \huge{.}}{ \blue{ \bold{ \: \: \: b= 3 }}}$

$\\ \: \: \: { \huge{.}}{ \blue{ \bold{ \: \: \: c= 2 }}}$

• So that –

$\\ \: \: \: { \huge{.}}{ \bold{ \: \: Sum \: \: of \: \: roots = A + B = - \frac{3}{8} }}$

$\\ \: \: \: { \huge{.}}{ \bold{ \: \: Product \: \: of \: \: roots = A . B = \frac{2}{8} }}$

• Now we have to find –

$\\ \: \: { \bold{ = { ( \dfrac{ {A}^{2} }{B}) }^{ \frac{1}{3} } + {( \dfrac{ {B}^{2} }{A}) }^{ \frac{1}{3} } }}$

$\\ \: \: { \bold{ = \: { \dfrac{ {A}^{ \frac{2}{3} } }{{B}^{ \frac{1}{3} } }} + { \dfrac{ {B}^{ \frac{2}{3} } }{A ^{ \frac{1}{3} } } } }}$

$\\ \: \: { \bold{ = \: { \dfrac{ {A}^{ \frac{2}{3} + \frac{1}{3} } + {B}^{ \frac{2}{3} + \frac{1}{3} } }{{B}^{ \frac{1}{3} }.A ^{ \frac{1}{3}}}} }}$

$\\ \: \: { \bold{ = \: { \dfrac{ {A}^{ 1 } + {B}^{ 1} }{(AB) ^{ \frac{1}{3}}}} }}$

$\\ \: \: { \bold{ = \: { \dfrac{ {A} + {B} }{(AB) ^{ \frac{1}{3}}}} }}$

• Now put the values –

$\\ \: \: { \bold{ = \: { \dfrac{ (- \frac{3}{8} ) }{( \frac{2}{8} ) ^{ \frac{1}{3}}}} }}$

$\\ \: \: { \bold{ = \: { \dfrac{ (- \frac{3}{8} ) }{ \frac{ {(2)}^{ \frac{1}{3} } }{2} }} }}$

$\\ \: \: { \bold{ = \dfrac{ - 3}{4 \times {2}^{ \frac{1}{3} } } }}$

$\\ \: \: { \bold{ = \dfrac{ - 3}{ {2}^{2} . {2}^{ \frac{1}{3} } } }}$

$\\ \: \: { \bold{ = \dfrac{ - 3}{ {2}^{ \frac{7}{3} } } }}$

Hence , $\: \: { \bold{ { ( \dfrac{ {A}^{2} }{B}) }^{ \frac{1}{3} } + {( \dfrac{ {B}^{2} }{A}) }^{ \frac{1}{3} } = \dfrac{ - 3}{ {2}^{ \frac{7}{3} } } }}$