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For the equation  \[a{x^4} + b{x^2} + c = 0\], then all the roots of the equation will be real if 
A. $b^2 < 4ac$
B. $b^2 >4ac$
C. \[b^2 = 4ac\]
D. $a=0, b=c$

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Last updated date: 23rd Jul 2024
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Answer
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Hint: We know that the nature of roots of any quadratic equation is totally decided by the term \[{b^2} - 4ac\]. We know that the roots can be real or imaginary and can be the same or distinct. This is dependent on \[{b^2} - 4ac\]. So let’s check for the given equation as they have mentioned that the roots are real. 

Complete step by step answer:
Given the equation is,  \[a{x^4} + b{x^2} + c = 0\].
Let \[{x^2} = t\]
Then the equation above becomes,
\[a{t^2} + bt + c = 0\]
Now this is the same as the general quadratic equation. Now we know that if,
\[\Rightarrow {b^2} - 4ac > 0\] then the roots are real and distinct.
\[\Rightarrow {b^2} - 4ac = 0\]  then the roots are real and the same.
\[\Rightarrow {b^2} - 4ac < 0\]   then the roots are imaginary and different.

To get Real roots, we have two cases:

Case 1: Roots are Real and Disctinct, if  $b^2-4ac>0$

$\Rightarrow b^2>4ac$

To satisfy this condition, we need a $b^2$ value which is greater than $4\times a \times c$ value.

Case 2: Roots are Real if $b^2-4ac=0$

$\Rightarrow b^2=4ac$

Therefore, If we observe the given options, Both (B) and (C) are correct.

 

Note: Here we have modified the given fourth degree equation into second degree equation to make it quadratic. The nature of roots can be found by the coefficients. 

Though the \[{b^2} - 4ac\] decides the nature of the roots. The formula \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]   gives the actual roots. 

If we check option (D), where coefficient $a$ is zero. It means it is not a quadratic equation.