# 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. $a=0, b=c$

**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:**

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.