1.

Find the value of a for which the zeros of alpha and beta of the quadratic polynomial x^2-6x+a=0 satisfy the relation 3alpha+2bita =20

Answer»

Given quadratic polynomial is x² - 6x + a = 0.

Here, a = 1, b = -6 and c = a

We have to find the value of a.

Now,

Here, alpha is denoted by p and beta by Q

Sum of ZEROS = -b/a

→ p + q = -(-6)/1

→ p + q = 6

Product of zeros = c/a

→ p × q = a/1

→ pq = a

Also given that, 3 alpha + 2 beta = 20

→ 3p + 2q = 20

→ (2p + p) + 2q = 20

→ 2p + 2q + p = 20

→ 2(p + q) + p = 20

From the above calculations, p + q = 6

SUBSTITUTE it in the above value

→ 2(6) + p = 20

→ 12 + p = 20

→ p = 20 - 12

→ p = 8

Substitute value of p in (p + q = 6)

→ 8 + q = 6

→ q = 6 - 8

→ q = -2

Now, we have p = 8 and q = -2. If we substitute their values in pq = a, then we can find the value of a.

→ 8(-2) = a

→ a = -16

Therefore, a = -16


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