Quadratics without real roots and their graphs

[Naveen Bhaskara]

Hi, so I studied that quadratics whose discriminants are less than 0 wouldn't have x-intercepts on the graph but I came across a problem and its solution which sort of contradicted this. Would someone please explain this?

 

[Dr.Peterson]

Hi, so I studied that quadratics whose discriminants are less than 0 wouldn't have x-intercepts on the graph but I came across a problem and its solution which sort of contradicted this. Would someone please explain this?View attachment 31754

I don't see a contradiction. Can you explain your thinking (and perhaps also show us what problem is being solved here)?

I suspect you may be confusing the quadratic equation in x, which you apparently want to have no real roots, and the quadratic equation in k, whose roots you need to find in order to solve the problem.

 

[pka]

I too sense a confusion in what you posted. You posted with the idea of non-real roots,
Such as HERE, you can see from the plot of the equation [imath]2x^2+3x+4=0[/imath] it has no real roots.
Moreover, that equation has two complex roots[imath]\quad-\dfrac{3}{4}+\dfrac{\sqrt{23}}{4}{\bf{i}}\quad \&\quad-\dfrac{3}{4}-\dfrac{\sqrt{23}}{4}{\bf{i}}[/imath]

Here is a fact that your mathematical background may not include:
note that the equation has real coefficients and the roots are complex conjugates
If [imath]a~\&~ b[/imath] are real numbers both [imath]z=a+b{\bf{i}}~\&~\overline{~z~}=a-b{\bf{i}}[/imath]
are both roots of [imath]t^2-2t+a^2+b^2=0[/imath] which is a quadratic with real coefficients, SEE HERE
Also SEE THIS.

[imath][/imath]

 

[Naveen Bhaskara]

Hi, thank you for the replies. Yes, I seem to have confused the equation of k with the equation of x. Sorry about that, I am new to the topic.
Again Thank you for the help.