Question

If alpha and bita are the zeroes of the polynomial xsq + x - 6 find the value of 1/alpha sq + 1/bita sq

Answer 1

Given :-

α and β are the zeroes of polynomial

x²+x-6

To find

$\sf \bigg(\dfrac{1}{\alpha}\bigg)^2+\bigg(\dfrac{1}{\beta}\bigg)^2$

A.T.Q:-

find the zeroes of polynomial

$\longrightarrow \sf x^2+x-6=0\\ \\ \longrightarrow\sf x^2 +(3-2)x-6=0\\ \\ \longrightarrow\sf x^2+3x-2x-6=0\\ \\ \longrightarrow\sf x(x+3)-2(x+3)=0\\ \\\longrightarrow\sf (x+3)(x-2)=0$

• Zeroes !

$\sf x+3=0 ,\ \ \ x =(-3)\\ \\\sf x-2=0 \ \ \ x=2$

• Then we have

$\boxed{\sf{\alpha= (-3)}}$

$\boxed{\sf{\beta= 2}}$

Now the value of $\sf \dfrac{1}{\alpha}+\dfrac{1}{\beta}$

$\longmapsto\sf \bigg(\dfrac{1}{\alpha}\bigg)^2= \dfrac{1}{-3}=\bigg(\dfrac{-1}{3}\bigg)^2\\ \\ \longmapsto\sf \dfrac{1}{\alpha^2}=\dfrac{1}{9}\\ \\ \longmapsto\sf \bigg(\dfrac{1}{\beta}\bigg)^2=\bigg(\dfrac{1}{2}\bigg)^2\\ \\ \longmapsto \dfrac{1}{\beta^2}=\dfrac{1}{4}\\ \\\longmapsto\sf \bigg(\dfrac{1}{\alpha}\bigg)^2+\bigg(\dfrac{1}{\beta}\bigg)^2\\ \\ \longmapsto\sf. \dfrac{1}{9}+\dfrac{1}{4}\\ \\ \longmapsto\sf \dfrac{ 4+9}{36}\\ \\ \longmapsto\sf \dfrac{13}{36}$

$\bigstar{\boxed{\sf{ \bigg(\dfrac{1}{\alpha}\bigg)^2+\bigg(\dfrac{1}{\beta}\bigg)^2=\dfrac{13}{36}}}}$

Answer 2

$\tt{\huge{\orange{Hello!!! }}} B.N.$

QUESTION :

If alpha and bita are the zeroes of the polynomial xsq + x - 6 find the value of 1/alpha sq + 1/bita sq

SOLUTION :

f(x) - : x^2 + x - 6

=> X^2 + 3 X - 2 x - 6

=> X( X + 3 ) - 2 ( X + 3 )

.=> ( X - 2 ) ( X + 3 )

=> The Zeroes, Alpha and Beta are 2 and -3 respectively...

=> 1/ { alpha}^2 = 1/ 4

=> 1 / { beta }^2 = 1/9

Sum => 1/4 + 1/9 = 13/36..........{A}