Question

If alpha and beta are the roots of x2-3x+1=0 then equation with roots alpha +1/beta and beta +1/alpha

Answer 1

Answer:

x2+3x+1

Step-by-step explanation:

let alpha be a and beta be b

a+b =-3

ab=1

a=1/b

b=1/a

sum =1/a+1/b=3/1

product =1/a×1/b=1/1

x2+3x+1

Answer 2

We need to recall the following rules for the roots of the quadratic equation.

If $\alpha ,\beta$ are the roots of a quadratic equation $ax^{2} +bx+c=0$ , then

• $\alpha +\beta =\frac{-b}{a}$
• $\alpha \beta =\frac{c}{a}$

This problem is about the roots of the quadratic equation.

Given:

$x^{2}-3x+1=0$

$\alpha ,\beta$ are the roots of the equation.

Then,

Sum of roots: $\alpha +\beta =-\frac{-3}{1}=3$                  .......$(1)$

Product of roots: $\alpha \beta =\frac{1}{1}=1$                      .......$(2)$

⇒ $\alpha =\frac{1}{\beta }$  or  $\beta =\frac{1}{\alpha }$                                      .......$(3)$

Thus, the sum of the roots of the required quadratic equation having roots $\alpha +\frac{1}{\beta }$ and $\beta +\frac{1}{\alpha }$  is,

$\alpha +\frac{1}{\beta }+\beta +\frac{1}{\alpha }$

$=\alpha +\alpha +\beta +\beta$                     .......From equation $(3)$

$=2\alpha +2\beta$

$=2(\alpha +\beta )$

$=2(3)$                                     .......From equation $(1)$

$=6$                                               .......$(4)$

The product of the roots of the required quadratic equation having roots $\alpha +\frac{1}{\beta }$ and $\beta +\frac{1}{\alpha }$  is,

$(\alpha +\frac{1}{\beta })*(\beta +\frac{1}{\alpha })$

$=(\alpha +\alpha)* (\beta +\beta)$                     .......From equation $(3)$

$=2\alpha *2\beta$

$=4\alpha\beta$

$=4(1)$                                          .......From equation $(1)$

$=4$                                                     .......$(5)$

Hence, the required quadratic equation is,

$x^{2} -$ sum of roots$(x)+$ product of roots$=0$

From equations $(4)$ and $(5)$ , we get

$x^{2} -6x+4=0$