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

Hey

Here is your answer,

α and β are the zeros of the polynomial x²-6x+a
∴, α+β=-(-6/1)=6 --------(1)
 αβ=a/1=a -----------(2)
3α+2β=20 -----------(3)

Multiplying (1) with 3 and (3) with 1 and subtracting we get,
3α+3β=18
3α+2β=20
∴, β=-2 

Putting in (1) we get, α=6-β=6+2=8
Then from (2), αβ=8(-2)=-16=a
∴, a=-16

Hope it helps you!

Answer 2

Step-by-step explanation:

x² - 6x + a

a = 1    b = -6   c = a

α +β = -b/a = 6

αβ = c/a = a

3α + 2β = 20

3α = 20 - 2β  

α = (20 - 2β) / 3

α + β = 6

(20 - 2β)/3 + β = 6

multiplying by 3 on both sides

20 - 2β + 3β = 18

20 + β = 18

β = 18 - 20  

β = -2

α = (20 -2β)/3

α = (20 - 2 × -2)/3

α = (20 + 4)/3

α = 8

αβ =  a  

8 × -2 = a

a = -16

Hope it helps!!

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