Question

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 1

Given polynomial

x²-6x + a = 0

To find:-

The value of 'a' for which the zeros α and β of the quadratic polynomial x²-6x+a=0 satisfy the relation 3α + 2β =20.

Solution:-

$\boxed{\alpha +\beta =\frac{-b}{a}}\\\boxed{\alpha \beta =\frac{c}{a}}$

α + β = -(-6)/1 = 6                ...(i)

αβ = a/1 = a                         ...(ii)

3α + 2β = 20

⇒2α + α + 2β = 20

⇒2(α + β) + α = 20

⇒2*6 + α = 20

⇒α = 20-12

⇒α = 8

Now use α = 8 in eq.(i):

α + β = 6

8 + β = 6

⇒β = -2

Now put the values of α and β in eq.(ii):

αβ = a

⇒8*-2 = a

⇒$\boxed{a = -16}$

∴Thus, the value of 'a' for which the zeros α and β of the quadratic polynomial x²-6x+a=0 satisfy the relation 3α + 2β =20 is (-16).

Answer 2

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