Question

If α and β are zeroes of the polynomial 3^2 + 2 + 3 ,then find 1/α^2 +1/β^2,urgent

Answer 1

Answer:

We have, f(x)=x2−1=0

We have, f(x)=x2−1=0⇒ (x−1)(x+1)=0

We have, f(x)=x2−1=0⇒ (x−1)(x+1)=0⇒ x=1 or −1

We have, f(x)=x2−1=0⇒ (x−1)(x+1)=0⇒ x=1 or −1⇒ α=−1,β=1

We have, f(x)=x2−1=0⇒ (x−1)(x+1)=0⇒ x=1 or −1⇒ α=−1,β=1New Roots are β2α and α2β

We have, f(x)=x2−1=0⇒ (x−1)(x+1)=0⇒ x=1 or −1⇒ α=−1,β=1New Roots are β2α and α2β=−2 or −2

We have, f(x)=x2−1=0⇒ (x−1)(x+1)=0⇒ x=1 or −1⇒ α=−1,β=1New Roots are β2α and α2β=−2 or −2⇒ Sum of new roots =−4

We have, f(x)=x2−1=0⇒ (x−1)(x+1)=0⇒ x=1 or −1⇒ α=−1,β=1New Roots are β2α and α2β=−2 or −2⇒ Sum of new roots =−4⇒ Product of new roots

We have, f(x)=x2−1=0⇒ (x−1)(x+1)=0⇒ x=1 or −1⇒ α=−1,β=1New Roots are β2α and α2β=−2 or −2⇒ Sum of new roots =−4⇒ Product of new roots

We have, f(x)=x2−1=0⇒ (x−1)(x+1)=0⇒ x=1 or −1⇒ α=−1,β=1New Roots are β2α and α2β=−2 or −2⇒ Sum of new roots =−4⇒ Product of new rootsa

Answer 2

Proper Question

If α and β are the two zeroes of the polynomial $3x^{2}+2x+3$, find $\dfrac{1}{\alpha ^{2}} +\dfrac{1}{\beta ^{2}}$.

Solution and Alternative Approaches

Solution

From the given polynomial we get the following.

$\implies \begin{cases} & \alpha +\beta =-\dfrac{2}{3} \\ & \alpha \beta =1 \end{cases}$

To find $\dfrac{1}{\alpha ^{2}} +\dfrac{1}{\beta ^{2}}$ we consider how to use the given information. If we add the two fractions, we get the following.

$\implies \dfrac{\alpha ^{2}+\beta ^{2}}{\alpha ^{2}\beta ^{2}}$

One way to find the value is using identity.

Consider the following two numbers.

➊ $\alpha ^{2}+\beta ^{2}$

➋ $\alpha ^{2}\beta ^{2}$

In the first number, we square both sides of the equation.

$\implies \alpha ^{2}+\beta ^{2}=(\alpha +\beta )^{2}-2\alpha \beta$

$\implies \alpha ^{2}+\beta ^{2}=(-\dfrac{2}{3} )^{2}-2$

$\implies \alpha ^{2}+\beta ^{2}=-\dfrac{14}{9}$

In the second number, we square both sides of the equation.

$\implies \alpha ^{2}\beta ^{2}=(\alpha \beta )^{2}$

$\implies \alpha ^{2}\beta ^{2}=1$

Hence, the required number is the following.

$\implies \dfrac{\alpha ^{2}+\beta ^{2}}{\alpha ^{2}\beta ^{2}}=\boxed{-\dfrac{14}{9}}$

Alternative Approach

Let the given polynomial be $P(x)=3x^{2}+2x+3$. Consider another equation having the inverse solutions of the polynomial. Such polynomial is $x^{2}\times P(\dfrac{1}{x} )$.

$\implies x^{2}\times P(x)=3x^{2}+2x+3$

Hence, we get the following.

$\implies \begin{cases} & \dfrac{1}{\alpha }+\dfrac{1}{\beta }=-\dfrac{2}{3} \\ & \dfrac{1}{\alpha \beta }=1 \end{cases}$

By squaring both sides of the equation, we get the following.

$\implies \dfrac{1}{\alpha ^{2}} +\dfrac{1}{\beta ^{2}} =(\dfrac{1}{\alpha } +\dfrac{1}{\beta } )^{2}-2\times \dfrac{1}{\alpha \beta }$

$\implies \dfrac{1}{\alpha ^{2}} +\dfrac{1}{\beta ^{2}} =(-\dfrac{2}{3} )^{2}-2$

$\implies \dfrac{1}{\alpha ^{2}} +\dfrac{1}{\beta ^{2}} =\boxed{-\dfrac{14}{9}}$