Question

If a and b are zeroes of polynomial p(x)=x2+x+1,then find the value of 1/a+1/b and a2+b2

Answer 1

Answer

The required values are :

1/a + 1/b = -1

a² + b² = -1

$\large\underline{\sf{Given}}$

The equation is

• x² + x + 1
• a and b are the zeroes of the given equation

$\sf\large\underline{Solution}$

We have from the sum relation of zeroes and coefficients :

$\sf Sum \ of \ the \ zeroes =- \dfrac{Coefficient \ of \ x}{Coefficient \ of \ x^{2}} \\\\ \sf\implies a + b = -\dfrac{1}{1} \\\\ \sf\implies a + b = -1 \longrightarrow (1)$

From the relation of product of zeroes and coefficients we have :

$\sf Product \ of \ zeroes =\dfrac{Constant \ term}{Coefficient \ of \ x^{2}} \\\\ \sf\implies ab = \dfrac{1}{1} \\\\ \sf\implies ab = 1 \longrightarrow (2)$

Dividing (1) by (2) we have :

$\sf\implies \dfrac{a+b}{ab}=\dfrac{-1}{1} \\\\ \sf\implies \dfrac{a}{ab} +\dfrac{b}{ab}=-1\\\\ \sf\implies \dfrac{1}{a}+\dfrac{1}{b} = -1$

Again squaring both side of (1) we have :

$\sf\implies (a+b)^{2} = (-1)^{2}\\\\ \sf\implies a^{2} +b^{2} +2ab = 1 \\\\ \sf\implies a^{2} + b^{2} + 2\times 1 = 1 \\\\ \sf\implies a^{2}+b^{2} = -1$