Question

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

Short answer

It is proved that the given equation has at most two real solutions.

Step-by-step solution

Rolle’s Theorem

Rolle’s theorem states that for any function \(f\left( x \right)\), there exists a number \(r\) in a domain \(\left( {a,b} \right)\) such that:

\({\rm{for r}} \in \left( {a,b} \right) \to f'\left( r \right) = 0\).

Solving the given function as per the theorem:

The given equation is:

\({x^4} + 4x + c = 0\)

Let,\(f\left( x \right) = {x^4} + 4x + c\),

We see the function is polynomial.

Then, it is differentiableas well as continuous for all values of \(x\).

Suppose the function is having three distinct real solutions\(a,\,\,b{\rm{ and c}}\,\,\left( {a < b < c} \right)\).

Then,

\(f\left( a \right) = f\left( b \right) = f\left( c \right) = 0\)

Now, according to Rolle’s Theorem, there must be two numbers \({r_1}{\rm{ and }}{r_2}\) such that:

\(f'\left( {{r_1}} \right) = f'\left( {{r_2}} \right) = 0\)

But,

\(\begin{aligned}{c}f'\left( x \right) &= 4{x^3} + 4 &= 0\\4\left( {{x^3} + 1} \right) &= 0\\4\left( {x + 1} \right)\left( {{x^2} - x + 1} \right) &= 0\\x &= - 1\end{aligned}\)

We have \(x = - 1\) as its only real solution.

This contradictsthe function is having three distinct real solutions.

Hence, there exist at most two real solutions for the given equation.