Question

If (a) and (ß) are zeroes of the polynomial x^2-2x-8,then form a quadratic polynomial whose zeroes are 2/a and 2/ß. (a=alpha,ß=beta)

Answer 1

Answer:

2x² + x - 1

Step-by-step explanation:

Polynomials in form of x^2 - Sx + P, represent S as sum of roots and P as product of roots.

Here, if a and ß are roots:

• a + ß = 2

• aß = - 8

Let the required polynomial be x² - px + q, and 2/a & 2/ß are roots, so

p = 2/a + 2/ß = 2(1/a + 1/ß)

p = 2(a + ß)/aß = 2(2)/(-8)

p = -1/2

Also,

q = (2/a)(2/ß)

q = 4(1/aß)

q = 4{1/(-8)} = - 4/8 = -1/2

Therefore, polynomial is:

x² - (-1/2)x + (-1/2)

(2x² + x - 1)/2

Required polynomial is 2x² + x - 1

Answer 2

Step-by-step explanation:

Given: (a) and (ß) are zeroes of the polynomial x² -2x - 8.

To find: A polynomial whose zeroes are 2/a and 2/ß.

Sum of zeros = -b/a

a + ß = 2/1

Product of zeros = c/a

aß = - 8/1

Now, assume that the polynomial is x² - Ax + B.

A = 2/a + 2/ß

A = 2(1/a + 1/ß)

A = 2(a + ß)/aß

Substitute the values,

A = 2(2)/(-8)

A = -1/2

Similarly,

B = (2/a)(2/ß)

B = (4/aß)

B = 4/(-8)

B = - 4/8

B = -1/2

Hence, the polynomial is x² - (-1/2)x + (-1/2) or x² + x/2 - 1/2.