Question

 If α and β are zeroes of the quadratic polynomial x2 – 6x + a; find the value of ‘a’ if 3α + 2β = 20.​

Answer 1

$\alpha + \beta = 6 \\ 3( \alpha + \beta ) = 18 \\ 3 \alpha + 2 \beta = 20 \\ \beta = - 2 \\ \alpha = 8 \\ \\ a = - 16$

Hope this answer is correct!!!!

(but I think that its incorrect).....

Answer 2

$\large\underline\pink{Given:-}$

• α and β are zeroes of the quadratic polynomial x2 – 6x + a
• 3α + 2β = 20

$\large\underline\pink{To find:-}$

• Fine the value of ‘a’ ....?

$\large\underline\pink{Solutions:-}$

• p (x) = x² - 6x + a
• 3α + 2β = 20 _______(i)

we know that:-

α + β = -b/a

α + β = -(-6)/1

α + β = 6

αβ = c/a

αβ = a/1

αβ = a

Now,

3α + 2β = 20

(3 + 2) (α + β) - 3β - 2α = 20

(3 + 2) (6) - 2α - 3β = 20

5 × 6 - 20 = 2α + 3β

30 - 20 = 2α + 3β

10 = 2α + 3β

2α + 3β = 10________(ii)

multiplying Eq. (i) by 2 and Eq. (ii) by 3, we get.

• 6α + 4β = 40 ________(iii)
• 6α + 9β = 30 _______(iv)

Subtracting Eq. (iii) from Eq. (iv).

${6α} \: + \: {4β} \: \: = \: \: {40} \\ {6α} \: + \: {9β} \: \: = \: \: {30} \\ \underline{ \: \: - \: \: \: \: \: \: \: \: - \: \: \: \: \: \: = \: \: \: - \: \: \: \: } \\ \: \: \: \: \: \: \: \: \: \: \: {-5β} \: \: \: \: \: = \: \: \: {10}$

$\: \: \: \: \: \leadsto \: \: β \: \: = \: \: {-2}$

putting the value β in Eq. (i)

3α + 2β = 20

3α + 2 (-2) = 20

3α - 4 = 20

3α = 20 + 4

3α = 24

α = 8

The value of α is 8 and β is -2.

Now, the value of a is

αβ = a

8 × (-2)

-16

Hence, the value of a is -16.