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Question

If a , b, c d ϵR, then the equation (x2+ax-3b)(x2-cx+b)(x2-dx+2b)=0 has

A
6 real roots
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B
at least 2 real roots
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C
4 real roots
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D
3 real roots
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Solution

The correct option is D at least 2 real roots
(x2+ax3b)(x2cx+b)(x2dx+2b)=0
For x2+ax3b=0
D=(a)24(1)(3b)=a2+12b
For x2cx+b=0D=(c)24(1)(b)=c24b
For x2dx+2b=0D=(d)2+4(1)(26)=d2+8b
Let us assume that x2+ax3b has imaginary y axis
then a2+12b<012b<a2b<a212
b is (-ve) then 4b willbe (+ve)
c2is(+ve)
So, c24b will be (+ve)
x2cx+b=0 has real roots
(x2+ax3b)(x2cx+b)(x2dx+2b)=0
Has 2 real roots.

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