2
$\begingroup$

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, $b^2-4ac \geqslant 0$

$x^2+(1-k)x-k=0$

$(1-k)^2-4*1*(-k)\geqslant0$

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

$1-2k+k^2+4k\geqslant0$

$k^2+2k+1\geqslant0$

$k^2+k+k+1\geqslant0$

$(k+1)(k+1)\geqslant0$

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?

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ Sorry, those "$k-1$"s should have been "$k+1$"s in my previous comment. $\endgroup$ – Barry Cipra Mar 7 '14 at 19:26
1
$\begingroup$

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 $k\in\mathbb{R}$.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
1
$\begingroup$

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 $k\in\mathbb{R}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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