Discriminants - When will a quadratic equation have 2 real solutions

1. Apr 4, 2006

danago

Hey. Ive got a question to solve, and im a bit confused.

Ok. I understand what the question is asking, i just dont know how to do it. Im oretty sure that if $\delta < 0$, thenthere are no real solutions, but im not sure about 1 or 2 solutions. I played around with quadratic functions on my calculator, and didnt really get anywhere.

I then need to do the same with the equation:
$$x^{2}+(k-3)x+9=0$$

And find the number of roots of the equation relative to the value of k.

If anyone could assist me, and guide me in the right direction, id appreciate it alot.

Thanks,
Dan.

EDIT: for some reason the LaTeX images arent showing up properly. Ill try to fix them.

Last edited by a moderator: Apr 8, 2006
2. Apr 4, 2006

Beam me down

The discriminant comes from the follwoing equation:

The discriminant is the bit in the square root. Now if you have a discriminant of <0 you end up with square roots of negative numbers. Which don't exist in the real number system.

If the discriminant equals 0, what does the equations shown in above simplify down to (hint: what is plus/minus the square root of 0)?

From that you should be able to see why there are two solutions when the discriminant is >0.

3. Apr 4, 2006

danago

ok that makes sense to me. Thanks alot for that.

The next part needs me to do the same with the equation:

x^2 + (k-3)x + 9 = 0

I found that:

if k=9, 1 solution
if k>9, 2 solutions
if k<9, no real solutions

Could someone check that and tell me if im correct?

Last edited: Apr 4, 2006
4. Apr 7, 2006

Beam me down

Ok the discriminant of x^2 + (k-3)x + 9 = 0 (eq. 1) is:

(k-3)^2-36
Which can be expanded to : k^2-6k-27 (eq. 2)

Equation 2's discriminant is >0, which indicates that equation 2 has two solutions for 0, which indicates there are two possible values for k in equation 1. You have found only one value (k=9) when there is only one solution, when there is in fact another solution for k (hint: its a negative integer).

5. Apr 7, 2006

danago

oh ok. I graphed it and found -3 to be the second solution. I didnt even think about a second solution at the time. Thanks for reminding me :)

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