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Question

# Which of the following statement is true about the equation x4+2x2âˆ’8x+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 rootsLet f(x)=x4+2x2−8x+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)2−8(−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)2−8(0)+3=3 f(1)=(1)4+2(1)2−8(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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