Question

if alpha and beta are zeros of the polynomial x square + 8x+6 from the polynomial whose zeros are (alpha+beta divided by alpha) + (alpha + beta divided by beta)​

Answer 1

Answer:p(x)= x²+8x+6

Here, a = 1, b = 8, c = 6

∵ ∝ and β are the zeroes of the polynomial

∴ we know,

∝ =  $\frac{-b}{a}$ = -8 and β =  $\frac{c}{a}$ = 6

given  , $\frac{\alpha +\beta }{\alpha }$ and  $\frac{\alpha+\beta }{\beta }$ are the zeroes of the required polynomial

Now,

⇒ Sum of zeros = $\frac{\alpha +\beta }{\alpha }+\frac{\alpha +\beta}{\beta }$

                            =  $\frac{(-8)+6}{-8} +\frac{-8+6}{6}$

                            = $\frac{-2}{-8} +\frac{-2}{6}$

                            = $\frac{1}{4} +\frac{-1}{3}$

                            = $\frac{3-4}{12}$  

                            = $\frac{-1}{12}$

product = $\frac{\alpha +\beta }{\alpha }*\frac{\alpha +\beta}{\beta }$

              =$\frac{(-8)+6}{-8} *\frac{-8+6}{6}$

              =$\frac{-2}{-8} *\frac{-2}{6}$

              =$\frac{4}{48}$

              =$\frac{1}{12}$

Required polynomial = k[$x^{2} -\frac{-1}{12}x+ \frac{1}{12}$]

= k[$\frac{12x^{2}+x+12 }{12}$ ]  

now, if  k = 12

∴ Required polynomial = 12x² + x + 12

Step-by-step explanation: