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Last updated date: 02nd Dec 2023
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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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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.

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.