Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For any real number $c$, the polynomial $x^3 + x + c$ has exactly one real root.

how can i show that above statement is true or false.can anyone help me please

share|improve this question

6 Answers 6

You can show (for example using the derivative) that the function $x \mapsto x^3 + x + c$ is increasing for any real $c$.

share|improve this answer
    
strictly increasing even. –  lhf Nov 23 '12 at 13:14

Here’s yet another argument, not using calculus. As @xen remarks, a strictly increasing real-valued function will have at most one root. Letting $f$ be our function, we find that $f(x+\varepsilon)-f(x)=\varepsilon\lbrack 3x^2+3x\varepsilon+\varepsilon^2+1\rbrack$, and the quantity in brackets is a sum of squares, namely $3x^2/4+(3x/2+\varepsilon)^2+1$. So the computed difference has the same sign as $\varepsilon$, the function is strictly increasing.

share|improve this answer

The easiest way is to show that $x \mapsto x^3+x+c$ is increasing or use Rolle's theorem. Another way is to use Descartes' rule of signs.
If $c>0$ it has a negative root while if $c\leq0$ it has a nonnegative root.

share|improve this answer

Hint: Apply Rolle's theorem.

See http://en.wikipedia.org/wiki/Rolle%27s_theorem

share|improve this answer

See wikipedia for a algebric condition in wikipedia section "Roots of a cubic function". Let $$ (1) \quad ax^3+bx^2+cx+d=0. $$ Every cubic equation $(1)$ with real coefficients has at least one solution ''x'' among the real numbers; this is a consequence of the intermediate value theorem. We can distinguish several possible cases using the discriminante $\Delta$ of $(1)$,

$$ \Delta = 18abcd -4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2. $$ The following cases need to be considered:

If $\Delta > 0$, then the equation has three distinct real roots.

If $\Delta = 0$, then the equation has multiple root and all its roots are real.

If $\Delta < 0$, then the equation has one real root and two nonreal complex conjugate roots.

For reference see the book

share|improve this answer
    
For a complet dedution of $\Delta$ see mathworld.wolfram.com/PolynomialDiscriminant.html and mathworld.wolfram.com/CubicFormula.html. I hope helped. –  Elias Nov 23 '12 at 13:34

$ x^3 + x + c = 0 $
$ x^3 + x = y = -c $
$ x(x^2 + 1) = y $

$ (x^2 + 1) $ is always positive and greater than zero, and $ x $ takes both negative and positive values. So it is possible to sweep the entire real numbers, including $ d $, by giving $ x $ values from $ -\infty $ to $ +\infty $. In other words, $ y=x(x^2 + 1) $ is a smooth function from $ \mathbb{R} $ to $ \mathbb{R} $.

graph of y=x(x^2 + 1)

So, $ x^3 + x + c = 0 $ must have at least real root, since $ y = x(x^2 + 1) $ always has a solution $x_0$ for any given $y_0$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.