Question

if alpha and beta are the zeros of quadratic polynomial x square - 2 x + 3 then find a quadratic polynomials whose zeros are alpha + 2, beta + 2​

Answer 1

Answer:

k[y² - 6x + 5] where k stands for any constant.

Step-by-step explanation:

If α and ß are zeros of polynomial,

→ x² - 2x - 3 = 0

→ x² - 3x + x - 3 = 0

→ (x + 1)(x - 3) = 0

→ x = - 1 and + 3

Hence, α = - 1 or + 3 and ß = + 3 or - 1

Now, α + 2 = - 1 + 2 = 1

Or 3 + 2 = 5

and ß + 2 = 3 + 2 = 5

Or - 1 + 2 = 1

In both cases final values are 5 and 1.

• p(y) = k[y² - (Sum of zeros)x + (Product of zeros)]

→ p(y) = k[y² - (5 + 1)x + (5 × 1)]

→ p(y) = k[y² - 6x + 5]

Answer 2

answer :

x^2-2x+3 is the given polynomial.

sum of zeroes -

alpha+beta = -b/a

= -(-2)/1

= 2

product of zeroes -

alpha.beta = c/a

= 3

according to the question.

2aplha/beta and 2beta/alpha are the zeroes of the polynomial and we have to find a polynomial by the given zeroes.

sum of the zeroes-

alpha+2+beta+2 = alpha+beta +4

= 2+4 (alpha+beta=2)

= 6

product of the zeroes -

(alpha+2) (beta+2) = alpha(beta+2) + 2(beta+2)

= alpha.beta + 2alpha + 2beta + 4

= alpha.beta + 2(alpha+beta) + 4

= 3+2(2)+4

(alpha.beta =3 , alpha+beta=2 )

= 3+4+4

= 11

Now, form a polynomial,

x^2 - (alpha+beta)x + alpha.beta

x^2 - (6)x + 11

x^2 - 6x + 11

Thank you.

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