Question

ʜᴇʏ ʜᴇʟᴘ ᴍᴇ ꜱᴏʟᴠᴇ ᴛʜɪꜱ Qᴜᴇ
ɢɪᴠᴇ ᴍᴇ ᴀɴꜱ ᴡɪᴛʜ ᴇxᴘʟᴀɴᴀᴛɪᴏɴ​

Answer 1

Answer :-2

Step-by-step explanation:

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 .

_________________________________

Answer 2

Answer:

2

is the right answer

Step-by-step explanation:

please mark as brainliest follow me please!