# Thread: Show that the equation , has no real roots for all real values of ݇k

1. ## Show that the equation , has no real roots for all real values of ݇k

Hi I am new here I hope I am posting in the right section

I am having some trouble with this sum I was hoping to get some help here.

2. ## Re: Show that the equation , has no real roots for all real values of ݇k

check the discriminant ...

$k^2 - 4(1)(k^2+2) = -(3k^2+8) < 0 \implies y = x^2+kx+(k^2+2)$ has no real zeros.

3. ## Re: Show that the equation , has no real roots for all real values of ݇k

may I ask how you got the numbers -4,1,3 and 8?

4. ## Re: Show that the equation , has no real roots for all real values of ݇k

Originally Posted by Taine145
may I ask how you got the numbers -4,1,3 and 8?
The discriminant is the part that goes inside the square root from the Quadratic Formula:

$x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

The discriminant is just $b^2-4ac$, which in the case of your quadtratic, you have $a=1, b=k, c=k^2+2$. Plug in, and see that for any real number $k$, you wind up with a negative discriminant (so when you take the square root, you get an complex number for the root).

5. ## Re: Show that the equation , has no real roots for all real values of ݇k

Another way- completing the square:
$\displaystyle x^2+ kx+ k^2+ 2= x^2+ kx+ \frac{k^2}{4}- \frac{k^2}{4}+ k^2+ 2= (x+ \frac{k}{2})^2- \frac{k^2}{4}+ \frac{4k^2}{4}+ \frac{8}{4}= (x+ \frac{k}{2})^2+ \frac{3k^2+ 8}{4}= 0$

That is the same as $\displaystyle (x+ \frac{k}{2})^2= -\frac{3k^2+ 8}{4}$. Since a square (of a real number) is never negative, that is possible only as long as [tex\-\frac{3k^2+ 8}{4}[/tex] is positive. But that in turn is true only if $\displaystyle 3k^2+ 8$ is negative which is not true for any k because $\displaystyle 3k^2$ is non-negative and 8 is positive.

6. ## Re: Show that the equation , has no real roots for all real values of ݇k

I am more familiar with this way, Thank you guys I really apprciate it

7. ## Re: Show that the equation , has no real roots for all real values of ݇k

As stated previously, $ax^2+bx+c = 0 \implies x = \dfrac{-b \pm \sqrt{\color{red}{b^2-4ac}}}{2a}$

The discriminant, $\color{red}{b^2-4ac}$, is a powerful tool to analyze quadratics ...

If $b^2-4ac < 0$, the quadratic has no real zeros

If $b^2-4ac = 0$, the quadratic has one real zero of multiplicity two

If $b^2-4ac > 0$, the quadratic has two distinct, real zeros ...

furthermore, if the discriminant is a perfect square value, the quadratic will factor over the rational numbers ... nice to know if you want to factor the quadratic.