If alpha and beta are the zeroes of the quadratic polynomial x2_ 6x+a ; find the value of 'a',if 3alpha+2beta =20
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α and β are the zeros of the polynomial x²-6x+a
∴, α+β=-(-6/1)=6 --------(1) and 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
∴, α+β=-(-6/1)=6 --------(1) and 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
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