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# 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: $$\frac {d} {dx} P(x) = 3x^2 + 3$$

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.

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