Find the range of values of $k$ for which the following equation $x^2+(1-k)x=k$ has real roots.

I tried it, for real roots, $b^2-4ac \geqslant 0$



${(1)^2-2\cdot1\cdot k+(k)^2}-4(-k)\geqslant0$





Solving the equation I get $k\geqslant -1$. Is this right?

In the book, the answer given is "all values of $k$"

What does that mean?

  • $\begingroup$ The book is correct. Show us how you got $k \geqslant -1$, and we can show you where you went wrong. $\endgroup$ – TonyK Mar 7 '14 at 19:07
  • $\begingroup$ What values did you use for $a$,$b$, and $c$? Perhaps you should show us more of what you did to get your answer. $\endgroup$ – Nate Mar 7 '14 at 19:09
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    $\begingroup$ You can use the Quadratic Formula. Easier, you can factor. $\endgroup$ – André Nicolas Mar 7 '14 at 19:09
  • $\begingroup$ @TonyK shown (edited) $\endgroup$ – Kiara Mar 7 '14 at 19:15
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    $\begingroup$ Sorry, those "$k-1$"s should have been "$k+1$"s in my previous comment. $\endgroup$ – Barry Cipra Mar 7 '14 at 19:26

The discriminant ($\Delta$) of the quadratic equation $x^2 + (1-k)x - k = 0$ is (using the $b^2 - 4ac$ "mnemonic") :

$$\begin{align}\Delta &= (1-k)^2 -4(1)(-k)\\ &=1 - 2k + k^2 + 4k\\ &= 1 + 2k + k^2\\ &= (k + 1)^2\end{align}$$

Now, a quadratic equation has only real roots if and only if $\Delta \ge 0$. This condition is met for all $k$, provided that $k \in \mathbb{R}$ (because the square of any real number is at least $0$).

It will generally fail otherwise (take $k = i = \sqrt{-1}$, the imaginary unit for example). It does not work for all numbers in general, if we were to consider complex and hypercomplex numbers.

So the required values of $k$ is $k\in\mathbb{R}$.

  • $\begingroup$ That means I should leave $(k+1)^2$ in this way and this is the answer? Sorry but can you be more simpler? @YiyuanLee $\endgroup$ – Kiara Mar 7 '14 at 19:41
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    $\begingroup$ Preferably, yes, because the square of any real number is non-negative (i.e. $\ge 0$. $\endgroup$ – Yiyuan Lee Mar 7 '14 at 19:47

Recall the formula $x_{1,2}=\frac{-b+-\sqrt{b^2-4ac}}{2a}$. This means that $x_{1,2}$ has real values if and only if $b^2-4ac$ is positive (because it is under square root). Now, since you've found that $b^2-4ac=(1+k)(1+k)=(1+k)^2\geqslant0$ it means that for any $k\in\mathbb{R}$ we can find $x_{1,2}$, so it has real roots for any $k\in\mathbb{R}$.


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