Question

Show that the equation \({x^4} + 4x + c = 0\) has at most two real roots.

Short answer

The given equation has at most two real roots.

Step-by-step solution

Given data

The equation is \({x^4} + 4x + c = 0\).

Concept of Rolle’s Theorem

“Let a function\(f\)satisfies the following conditions,

1. A function\(f\)is continuous on the closed interval\((a\;,\;b)\).

2. A function\(f\)is differentiable on the open interval\((a\;,\;b)\).

3.\(f(a) = f(b)\)

Then, there is a number\(c\)in open interval\((a\;,\;b)\)such that\({f^\prime }(c) = 0\).”

Calculation to check the given equation has at most two real roots

Suppose \(f(x)\) has three distinct real roots \(a,b\) and \(r\) where \(a < b < r\).

Then, \(f(a) = f(b) = f(r) = 0\).

Since the polynomial is continuous on the closed interval \((a,b)\) and \((b,r)\) and differentiable on the open interval \((a,b)\) and \((b,r)\), then by Rolle's Theorem, there exist a number \({c_1}\) and \({c_2}\) with \(a < {c_1} < b\) and \(a < {c_2} < r\) in \((a,b)\) and \((b,r)\) respectively such that \({f^\prime }\left( {{c_1}} \right) = {f^\prime }\left( {{c_2}} \right) = 0\).

So, \({f^\prime }(x) = 0\) must have at least two real solutions.

Obtain the derivative of \(f(x)\).

\(\begin{aligned}{c}{f^\prime }(x) &= \frac{d}{{dx}}\left( {{x^4} + 4x + c} \right)\\{f^\prime }(x) &= 4{x^3} + 4\\{f^\prime }(x) &= 4\left( {{x^3} + 1} \right)\end{aligned}\)

Equate the above equation to 0 to get the roots of the derivate of the given function \(f\).

\(\begin{aligned}{c}4\left( {{x^3} + 1} \right) &= 0\\{x^3} + 1 &= 0\\{x^3} &= - 1\\x &= \sqrt(3){{ - 1}}\end{aligned}\)

Thus, the derivative of \(f(x)\) has only one real solution \(x = - 1\) this contradicts the fact that \({f^\prime }(x) = 0\) must have at least two real solutions.

Thus, it can be concluded that the given equation cannot have three real roots.

Hence, the given equation has at most two real roots.