Question

if alpha and betta are the zeroes of the polynomials f(x)= x^2-4x-5 ,then find the value of alpha^2+ betta^2 ..??​

Answer 1

Answer:

10x² - 9x + 2 .l

Given:

Alpha and betta are the zeroes of 2x²-9x+10,

To form:

Polynomial whose zeroes are 1/alpha and 1/beta

Solution:

First we should know the value of alpha and beta .

\begin{gathered}2 {x}^{2} - 9x + 10 = 0 \\ \\ 2 {x}^{2} - 5x - 4x + 10 = 0 \\ \\ x(2x - 5) - 2(2x - 5) = 0 \\ \\ (x - 2)(2x - 5)\end{gathered}

2x

2

−9x+10=0

2x

2

−5x−4x+10=0

x(2x−5)−2(2x−5)=0

(x−2)(2x−5)

Either, x - 2 = 0

x = 2

or, 2x - 5 = 0

or, 2x = 5

or, x = 5/2

So, Alpha = 5/2 or 2

Beta = 2 or 5/2

Hence, 1/Alpha = 2/5 or 1/2

1/Beta = 1/2 or 2/5

Therefore the polynomial whose zero is 1/alpha and 1/beta is ......

x² - (sum of zeros)x + product of zeros

x² - (2/5 + 1/2)x + (2/5×1/2)

x² - (4+5)/10x + 1/5

x² - 9x/10 +1/5

And in balanced form,

10x² - 9x + 2

The required polynomial is 10x² - 9x + 2 .

I hope it will help you