Question

If \[{{b}^{2}}\ge 4ac\], for the equation \[a{{x}^{4}}+b{{x}^{2}}+c=0\], then all the roots of the equation will be:
(This question has multiple correct options)
(a) b > 0, a < 0, c > 0
(b) b < 0, a > 0, c > 0
(c) b > 0, a > 0, c > 0
(d) b > 0, a < 0, c < 0

Answer 1

Hint: First of all consider the biquadratic equation and in that substitute \[{{x}^{2}}=t\] and make a quadratic equation \[a{{t}^{2}}+bt+c\]. As we are given that \[{{b}^{2}}\ge 4ac\], so this equation would have real roots. Now, as \[{{x}^{2}}=t,\text{ so }x=\pm \sqrt{t}\]. So, now for real x, t must be positive. So, take the roots of the quadratic equation positive and find the sign of a, b, and c by using the sum and product of the roots which are equal to \[\dfrac{-b}{a}\text{ and }\dfrac{c}{a}\] respectively.
Complete step-by-step answer:
We are given that \[{{b}^{2}}\ge 4ac\], for the equation \[a{{x}^{4}}+b{{x}^{2}}+c=0\]. We have to find the condition for which the roots of the equation will be real. Let us consider the equation given in the question.
\[a{{x}^{4}}+b{{x}^{2}}+c=0.....\left( i \right)\]
We can also write the above equation as,
\[a{{\left( {{x}^{2}} \right)}^{2}}+b{{x}^{2}}+c=0\]
By substituting \[{{x}^{2}}=t\] in the above equation, we get,
\[a{{t}^{2}}+bt+c=0...\left( ii \right)\]
We can see that the above equation is a quadratic equation because its degree is 2. We know that for a quadratic equation to have real roots, \[\text{Discriminant }\ge \text{0,i}\text{.e}\text{., }{{b}^{2}}-4ac\ge 0\].
\[{{b}^{2}}\ge 4ac\]
We are already given the above equation in the question. That means equation (ii) will have real roots as the value of t would be real.
Now, as we know that,
\[{{x}^{2}}=t\]
\[x=\pm \sqrt{t}....\left( iii \right)\]
From the above equation, we can clearly see that if we want the roots of the equation (i) to be real or x to be real, t must be positive otherwise x would be an imaginary number. So, now let us find the condition when ‘t’ that is the roots of equation (ii) would be positive. We know that,
\[\text{Sum of roots }=\dfrac{-\left( \text{coefficient of x} \right)}{\left( \text{coefficient of }{{\text{x}}^{2}} \right)}\]
So, we get the sum of the roots \[=\dfrac{-b}{a}\]
We know that the roots of equation (ii) are positive. So, we get the sum of the roots \[=\dfrac{-b}{a}>0\]
So, we get,
\[\dfrac{b}{a}<0....\left( iv \right)\]
Also, we know that,
\[\text{Product of roots }=\dfrac{+\left( \text{constant term} \right)}{\left( \text{coefficient of }{{\text{x}}^{2}} \right)}\]
So, we get the product of roots \[=\dfrac{c}{a}\]
We know that the roots of equation (ii) are positive. So, we get the product of roots
\[\dfrac{c}{a}>0....\left( v \right)\]
In equation (iv), we can see that \[\dfrac{b}{a}<0\]. That means b and a would have opposite signs that if b > 0, then a < 0….(vi) and if b < 0, then a > 0…..(vii).
In equation (v), we can see that \[\dfrac{c}{a}>0\]. That means c and a would have the same sign that is if, c > 0, then a > 0 …..(viii) and if c < 0, then a < 0…..(ix).
By combining equation (vi) and (ix), we get, b > 0, a < 0 and c < 0.
By combining equation (vii) and (viii), we get, b < 0, a > 0 and c > 0.
Hence, option (b) and (d) are the right answers.

Note: In this question, many students make this mistake of taking just one option out of (b) and (d) and leaving the other, this must be taken care of. Also, in this question, some students think that the given information that is \[{{b}^{2}}\ge 4ac\] is irrelevant which is the wrong opinion because before considering t as positive, we must ensure that t should be real as well, then x can never be real even if t would be positive. So, students must be clear about the concepts.