Question

if alpha and beta are the zeros of the polynomial x2+4x+3 find the values for 1/alpha + 1/beta AND 1/ALPHA SQUARE +1/beta square

Answer 1

this is the answer for your question.

Answer 2

The value of $\bold{\frac{1}{\alpha}+\frac{1}{\beta}=1 \frac{1}{3}}$

The value of $\bold{\frac{1}{\alpha}+\frac{1}{\beta}=1 \frac{1}{9}}$ .

Solution:

The polynomial equation is  $x^{2}+4 x+3=0$,  

To find the roots of the equation we use the given polynomial equation

$x^{2}+3 x+x+3=0$

x(x+3)+1(x+3)=0  

(x+3)(x+1)=0    

x=3 and x=1    

$\bold{\alpha=3, \beta=1}$

To find the value $\bold{\frac{1}{\alpha}+\frac{1}{\beta}}$:

Substituting the value of $\alpha$ and $\beta$ in $\frac{1}{\alpha}+\frac{1}{\beta}$

$\frac{1}{\alpha}+\frac{1}{\beta}=\frac{1}{3}+\frac{1}{1}=1 \frac{1}{3} \\\\Hence, \frac{1}{\alpha}+\frac{1}{\beta}=1 \frac{1}{3}$

We put $\alpha=3, \beta=1$

To find the value $\bold{\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}}$ :

Substituting the value of $\alpha \text { and } \beta \text { in } \frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}$

$\left(\frac{1}{\alpha}\right)^{2}+\left(\frac{1}{\beta}\right)^{2}=\left(\frac{1}{3}\right)^{2}+\left(\frac{1}{1}\right)^{2}=1 \frac{1}{9}$

Hence, $\frac{1}{\alpha}+\frac{1}{\beta} \text { is } 1 \frac{1}{9}$ .