Question

If alpha and beta are zeros of polynomial x2+8x+6 form a polynomial whose zeros are 1+beta/alpha and 1+alpha/beta

Answer 1

Hello,

The relation between the zeroes and coefficients of a quadratic equation are as follows:
$\alpha + \beta = \frac{ - b}{a} \\ \alpha \beta = \frac{c}{a}$
For every polynomial of form ax²+bx+c

Now the given polynomial is
P(x) =x²+8x+6
So for this polynomial
$\alpha + \beta = - 8 \\ \alpha \beta = 6$
Now,
We are asked a polynomial whose zeroes are
$\frac{1 + \beta }{ \alpha } \: \: \: and \: \: \frac{1 + \alpha }{ \beta }$
Now let the polynomial be f(x)
So,
For f(x) the sum of the zeroes is
$\frac{1 + \beta }{ \alpha } + \frac{1 + \alpha }{ \beta } = \frac{ \beta (1 + \beta ) + \alpha (1 + \alpha )}{ \alpha \beta } \\ which \: is \: equal \: to \\ \frac{ \beta + { \beta }^{2} + \alpha + { \alpha }^{2} }{ \alpha \beta } \\ = \frac{ \alpha + \beta + ( { \alpha + \beta )}^{2} - 2 \alpha \beta }{ \alpha \beta } \\ = \frac{ - 8 + {( - 8)}^{2} - 2 \times 6}{6} \\ = \frac{ - 20 + 64}{6} = \frac{44}{6}$

Now, for the product of zeroes of f(x)
$\frac{1 + \beta }{ \alpha } \times \frac{1 + \alpha }{ \beta } \\ = \frac{(1 + \beta )(1 + \alpha )}{ \alpha \beta } \\ = \frac{1 + \alpha + \beta + \alpha \beta }{ \alpha \beta } \\ = \frac{1 - 8 + 6}{6} = \frac{ - 1}{6}$
Now,
For f(x)
Sum of roots = 44/6
Product of zeroes = - 1/6

The structure of the polynomial is
K(x²-(sum of zeroes) x +(product of zeroes)

So, f(x) = k(x²-44/6-1/6)
Or, f(x) = 6x²-44x-1
When k = 6

Hope this will be helping you....

Answer 2

Zeroes = alpha and beta

Sum of zeroes = Alpha + beta = -b/a = -8

Product of zeroes = Alpha * beta = c/a = 6.

Then, for futher steps, refer the attachment above... !

Hope this helps!