TIFR 2013 problem 3 | Roots of an equation

Try this beautiful problem from TIFR 2013 problem 3 based on the roots of an equation.

Problem: TIFR 2013 problem 3 

The equation \(x^3+3x-4=0\) has exactly one real root.

Discussion

Hint: Try differentiating. Is the function monotonic?

Note that the cubic equation has at least one real root (fundamental theorem of algebra provides for three roots counting repetitions, real or complex; for equations with real coefficients, the complex roots appear in conjugate pairs. Hence a cubic equation with real coefficients must have at least one real root).

Now differentiating \( P(x) = x^3 + 3x - 4 \) we have the following: $$ \frac {d} {dx} P(x)  = 3x^2 + 3 $$

What can you say about this derivative?

Clearly it is always positive (perfect square plus three is at least three or more).

Hence P(x) is monotonically increasing. It will 'cut' the x-axis only at one point.

Therefore P(x) has exactly one real root.

Remark:

In this problem, first we established P(x) has at least one real root. Then we proved it has exactly one.  Both steps are crucial.

Some Useful Links:

• What is TIFR and how to prepare for it?
• TIFR 2014 Problem 22
• TIFR - Problem Garden
• Fundamental Theorem of Abelian Groups – Video