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Question

Which of the following statement is true about the equation x4+2x28x+3=0 ?

A
There are 2 positive real roots, 2 imaginary roots
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B
There are 4 imaginary roots
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C
There are 2 negative real roots
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D
There are 1 positive real root, 1 negative real root, 2 imaginary roots
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Solution

The correct option is A There are 2 positive real roots, 2 imaginary roots
Let f(x)=x4+2x28x+3.
According to the Fundamental Theorem of Algebra, a n degree polynomial equation should have n roots.
Thus, f(x) should also have 4 roots (real or imaginary).
Using Descartes Rule of Signs,
The number of sign changes in f(x) is 2
Maximum number of positive real roots =2.
Possible number of positive real roots =2 or 0.

& f(x)=(x)4+2(x)28(x)+3
f(x)=x4+2x2+8x+3
Thus, the number of sign changes in f(x) is 0
Maximum number of negative real roots =0
Possible number of negative real roots =0

From the above observations, the different possibilities are given below.

Now, let's try to find the value of f(x) at different points
f(0)=(0)4+2(0)28(0)+3=3
f(1)=(1)4+2(1)28(1)+3=2
Here, f(0)>0 and f(1)<0
f(x) cuts x axis at least once between 0 and 1.
The given equation has at least one real root.
Possibility 1 is valid.

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