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Question

The number of real roots of the equation 3x4+25x3+6x2+25x+3=0 is

A
1
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B
0
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C
4
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D
2
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Solution

The correct option is D 2
Given equation is 3x4+25x3+6x2+25x+3=0
This is a type 1 biquadratic equation.
On dividing by x2 and rearranging we get,
3(x2+1x2)+25(x+1x)+6=0
Substituting, x+1x=t, we get:
x2+1x2=t22
We can write the above equation as,
3(t22)+25t+6=0
3t2+25t=0
t(3t+25)=0
t=253,0
But t0 as t(,2][2,)

x+1x=253

3x2+3=25x
3x2+25x+3=0
Using discriminant method to solve this equation we get,
x=25±625366
x=25±5896
Hence, there are 2 real roots of the equation.

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