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Answer :-2
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Let the root of 1st and 2nd equation is (α, 4β) and (α, 3β)
given 1st equation x²-6x+a = 0
products of the root is 4αβ=a ______(1)
Given 2nd equation x²-cx+6=0
Product of the root is 3αβ= 6_____(2)
αβ = 2
put value of αβ =2 in (1)
4αβ = a =8 from (1)
∴ roots of the equation x²-6x +8
x² -6x +8 => x²-4x -2x +8 =0
=> x(x-4)-2(x-4) =0
=>(x-4)(x-2) =0
hence roots of the equation x²-6x +8 is (2, 4 )
So , the value of α can be (2, 4)
but given in question that , other roots , 4β, 3β are integers .
in this case , if we put value of α =2 in (2)
3αβ = 6 => αβ = 2 it satisfy
in case , if we put value of α = 4 in (2)
3αβ = 6 => αβ = 1/2 that is not an integer .
Since , common root α = 2 Answer .
_________________________________
Answered by
1
Answer:
2
is the right answer
Step-by-step explanation:
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