Question

If the value of a discriminant is 13, what does that tell us about the type and number of roots?

1 rational root

2 rational roots

2 complex conjugate roots

2 irrational roots

Answer 1

What do you know about the discriminant?

Answer 2

@ladyb What do you think?

Answer 3

Your solutions are given by: \[\Large x=\frac{ -b\pm \sqrt{b^2-4ac} }{ 2a}\]

And the discriminant is the \(\Large b^2-4ac\) part of that, right?

Answer 4

So think about what kind of number you will get from the quadratic equation, if the \(\Large b^2-4ac\) part is =13.

Answer 5

In fact, think about what WOULD have to be true of the discriminant, for each of these cases:

1 rational root?
2 rational roots?
2 complex conjugate roots?
2 irrational roots?

In each case, you can describe under what conditions on the discriminant, you will get the type of roots described. :)

Answer 6

okay @ Debbie G im unsure how to solve this , can you walk me through the steps to solve it

Answer 7

Do you know what the discriminant is?

Do you know what the quadratic formula is? (I gave it above, it's the x= part).

Tell me what that I posted above, you are unclear on, and we can go from there.

Answer 8

no i dont no the quadratic formula @ Debbie G

Answer 9

Well, you are obviously studying quadratic equations in this class (what is the class?), given that you are asked this problem. Does this look familiar?

\[\Large x=\frac{ -b\pm \sqrt{b^2-4ac} }{ 2a}\]That is the equation for the roots - also called the "zeros" - of a quadratic equation, which is an equation of the form\[\Large ax^2+bx+c=0\]

Does any of that ring a bell?

Answer 10

In other words, when you have an equation that looks like this: \(\Large ax^2+bx+c=0\)

And you want to SOLVE it (that is, find the values for x that make it true), you plug the a, b, and c into this:

\[\Large x=\frac{ -b\pm \sqrt{b^2-4ac} }{ 2a}\]

And that gives you the x's that solve it. You might get 1 rational solution, 2 rational solutions, 2 irrational solutions, or 2 complex solutions, depending on the stuff that is under the radical sign, the \(b^2-4ac\), which is called the discriminant.

Answer 11

yes because i remember a,b, and c , can have any value except that a cant be 0.

Answer 12

Right!! Good! The only reason we don't want a=0 is because then our \(x^2\) term would go away, and it isn't a quadratic equation anymore, it's just a plain ol' linear equation (which is perfectly fine, but a different type of equation.) :)

Answer 13

Now when you look at \(\Large x=\dfrac{ -b\pm \sqrt{b^2-4ac} }{ 2a}\) keep in mind, you can also write that as \(\Large x=\dfrac{ -b }{ 2a} \pm\dfrac{ \sqrt{b^2-4ac} }{ 2a}\)

Answer 14

okay i get that part but now what i do now that i know the Quadratic formula

Answer 15

hold one sec....

Answer 16

ok

Answer 17

ok, so you will get different KINDS of numbers from that, depending on what kind of number \(b^2-4ac\) is.

For example, if \(b^2-4ac\) =0, then it reduces to just the single solution, \(\dfrac{-b}{2a}\) , which is a rational number (it has to be, since a, b, and c are all integers).

So I've just given you a hint: when \(b^2-4ac\) =0 there is one rational root. Does that make sense?

Answer 18

no i still dont get it

Answer 19

if \(\Large b^2-4ac=0\)

then \(\Large x=\dfrac{ -b }{ 2a} \pm\dfrac{ \sqrt{0} }{ 2a}=\dfrac{ -b }{ 2a} \)

which is ONE RATIONAL NUMBER. It has to be rational because a and b are integers. It is only ONE solution because the 0 discriminant means that the \(\pm\) part of the quadratic formula goes away. So

\(\Large \text{discriminant}=b^2-4ac=0\) guarantees ONE RATIONAL SOLUTION.

Tell me if that makes sense, and if not, what don't you understand before we go on?

Answer 20

okay now i get it

Answer 21

Good. The whole problem, or just what I wrote above? Do you want me to continue?

Answer 22

i just understand what u wrote above

Answer 23

Now looking at \(b^2-4ac\), under what circumstances - what KIND of values for \(b^2-4ac\), would
\(\Large x=\dfrac{ -b }{ 2a} \pm\dfrac{ \sqrt{b^2-4ac} }{ 2a}\) give you:

2 rational roots?
2 complex conjugate roots?
2 irrational roots?

What would have to be true of \(b^2-4ac\)? We see why when \(b^2-4ac=0\), we only got ONE root, so no need to worry about that case. For any other (non-zero) value of \(b^2-4ac\), we will certainly get 2 results, because of the \(\pm\). So all we really need to worry about then is, what would be true of \(b^2-4ac\) in order that

\(\Large \dfrac{ \sqrt{b^2-4ac} }{ 2a}\) would be rational? would be complex? would be irrational?

Answer 24

Start with complex. That's probably the easiest one. :) What must be true of \(b^2-4ac\) to make that expression above COMPLEX?

Answer 25

i am un sure :(

Answer 26

Do you know what a complex number is?

Answer 27

A complex number is a number that can be expressed in the form a + bi, where a and b are real numbers and i is the imaginary unit, where i2 = −1.

Answer 28

ok, good, you know how to google. ;) lol.... now, technically, ALL REAL numbers are also complex numbers (but not all complex numbers are real numbers), but for our purposes, I'm just saying "complex number" to mean a number that has an "i" in it.

So how does that "i" get there? It gets there when we try to take a square root of a NEGATIVE number, right?

\(\sqrt{-9}=i\sqrt{9}=3i \)

Answer 29

So what has to be true about \(\sqrt{b^2-4ac}\)

in order that \(\Large \dfrac{ \sqrt{b^2-4ac} }{ 2a}\) would be COMPLEX?

Answer 30

you would have to make 2a 2i right?

Answer 31

no....

Answer 32

oh im confuse

Answer 33

The 2a is 2a. It is, what it is. You get a complex number when you take a SQUARE ROOT of a NEGATIVE NUMBER.

So \(\Large \dfrac{ \sqrt{b^2-4ac} }{ 2a}\) will be complex, if \(\Large b^2-4ac\) IS WHAT??

Answer 34

if it were negative then it would be b^2 - 4aci , correct or am i way off ?

Answer 35

well, you're getting there. :)

If \(\Large b^2-4ac<0\) then \(\Large \sqrt{b^2-4ac}\) is a complex number, because you have something NEGATIVE under that square root. So that means that

\(\Large x=\dfrac{ -b }{ 2a} \pm\dfrac{ \sqrt{b^2-4ac} }{ 2a}\) gives us 2 COMPLEX SOLUTIONS which are the "complex conjugates".

So now you know when you get 1 rational solution, and when you get 2 complex conjugate solutions.

And neither case happens when the discriminant, \(\Large b^2-4ac=13\). So what's left is, what does it take to get:

2 rational roots?
2 irrational roots?

Answer 36

We know that it isn't \(\Large b^2-4ac\le0\) because we've covered those cases above. so we know \(\Large b^2-4ac>0\), which by the way, is true for \(\Large b^2-4ac=13\). So the only question is, what about \(\Large b^2-4ac\) makes the two solutions RATIONAL NUMBERS vs. IRRATIONAL NUMBERS??

Well, a rational number is just a number of the form \(\Large \dfrac { m }{n}\)where m and n are integers. So how would \(\Large \dfrac{ \sqrt{b^2-4ac} }{ 2a}\) be an integer? We already know that 2a is an integer, so we just need

\(\Large \sqrt{b^2-4ac}\) to be an integer, which it will be if the radical part simplifies to an integer, which it will if the stuff under the square root (THE DISCRIMINANT) is a perfect square. If the stuff under the square root is NOT a perfect square, then you can't get rid of that square root, so you are left with an IRRATIONAL NUMBER.

For example:
\(\Large \sqrt{16}=4\) so if \(\Large b^2-4ac=16\) (or any perfect square) then you will have 2 rational roots.

But \(\Large \sqrt{20}=\sqrt{4\cdot5}=2\sqrt{5}\), so if \(\Large b^2-4ac=20\) (or any positive number that is NOT a perfect square, like say 13) then you will have 2 IRRATIONAL roots.

Answer 37

@ladyb I know I threw a lot of stuff at you in this thread. I could have just given you the answer 2 hours ago, but hopefully you learned something from our discussion and have gained a better understanding of quadratic functions and their roots. If so, you will have an easier time answering a question like this on your own, next time. :)

Answer 38

i appreciate you for teaching me and not just giving me the answer i now have a better understanding of quadratic equations :)