Question

Find the value of the discriminant and describe the number and type of roots.
3x2 - 8x + 2 = 0

Answer 1

@tkhunny

Answer 2

\(b^{2} - 4ac\)? This should ring bells. Have you used the quadratic formula?

Answer 3

no this is new to me

Answer 4

You've seriously never used the quadratic formula? Just checking.

Answer 5

no never

Answer 6

(sigh) Why are you given problem that you have no tools to solve?

Have you ever solved a quadratic equation by "Completing the Square"?

Answer 7

no

Answer 8

Gaaa!! Who writes these course materials?!

Okay, a GENERAL quadratic in STARNDARD form is this \(ax^{2} + bx + c = 0\), where a, b, and c are Real Numbers. Do you believe?

Answer 9

yes

Answer 10

One day, you should learn the process called "Completing the Square". I guess we'll have to wait for another lesson on another day for that. Suffice it to say that if you use this technique to SOLVE this quadratic equation, you will get:

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

Since we can't help you prove that, as you haven't the tools, you'll jus have to believe me. a, b, and c in the formula are the same as we just saw in the Standard Form.

Following?

Answer 11

yes

Answer 12

This will sound like a silly question. What value for 'b' will make this formula fail?

Answer 13

idk

Answer 14

Actually, that's an excellent answer. There is no such value. 'b' is just 'b'. It tells us nothing by itself.

Okay, what value for 'a' can make this formula fail? Think carefully! "idk" is not acceptable on this one.

Answer 15

nothing

Answer 16

?? If by "nothing" you mean a = 0 is bad, then you are exactly correct. You can't have zero (0) in the denominator, can you?

Answer 17

no you cant have 0 in a denominator

Answer 18

Great, so a better not be zero. A review of the standard form

ax^2 + bx + c = 0 suggests for a = 0, it isn't even a quadratic equation! So, a better not be zero - ever!

One more silly question. What sorts of values can you put in a square root?

\(\sqrt{4}\) -- Does that work?
\(\sqrt{0}\) -- Does that work?
\(\sqrt{-3}\) -- Does that work?

Answer 19

4 yes but the rest is no cus nothing multiplies by itself to get that particular number such as 0 and -3

Answer 20

Careful. 0*0 = 0. That one works.

What we are saying, for now, is that a square root can handle anything that is NOT NEGATIVE. Agreed?

Answer 21

yes

Answer 22

Okay, we are almost there. Tired yet?

Way back at the beginning, I wrote the quadratic formula. We talked about the 'b' out in front. We talked about the 'a' in the denominator. The only part we have not talked about is the square root.

\(\sqrt{b^{2} - 4ac}\)

Since we just decided that square roots can't handle anything negative, if we get \(b^{2} - 4ac < 0\) we are just totally out of luck if we want to find a Real Number. Does this make sense?

Answer 23

yes kinda

Answer 24

Suppose we have \(2x^{2} + 3x + 4 = 0\)

a = 2 -- We are very happy this is not zero.
b = 3
c = 4

\(b^{2} - 4ac = 3^{2} - 4(2)(4) = 9 - 32 = -23\)

If we are to find a solution, we will need to evaluate \(\sqrt{-23}\).

Since we don't like \(\sqrt{-23}\), we throw our hands in the air and exclaim, "There is no solution in the Real Numbers!"

How are we doing?

Answer 25

great im following

Answer 26

Well, that's what happens when \(b^{2} - 4ac < 0\)

What do you suppose we get to think if \(b^{2} - 4ac > 0\)?

Answer 27

9-32>0
+32 +32
9>32
-9
\[\sqrt{23}\]
yes i think

Answer 28

?? I don't know what you did? 9 - 32 will always be -23. There is nothing you can do to change that.

Suppose we have \(2x^{2} +3x - 4 = 0\)

a = 2 -- We are very happy this is not zero.
b = 3
c = -4

\(b^{2} −4ac=3^{2} −4(2)(-4)=9+32= 41 \)

If we are to find a solution, we will need to evaluate \(\sqrt{41}\).

Can we do that?

Answer 29

ohhh i see what i did wrong

Answer 30

Good. How about \(\sqrt{41}\)? Is that something that exists? Are we happy with that result?

Answer 31

yes

Answer 32

Perfect. Now, this one has a special case. 41 is pretty ugly sitting under a square root. What do you suppose we can say if \(b^{2} - 4ac\) is a Perfect Square?

Side Question: Have you solved a quadratic equation by FACTORING?

Answer 33

yes

Answer 34

idk

Answer 35

That's good. We'll need that.

Take a look at that Quadratic Formula. If \(b^{2} - 4ac\) is a perfect square, we can just plain not worry about the square root.

Suppose we have \(x^{2} + 3x − 4 = 0\)

a = 1 -- We are very happy this is not zero.
b = 3
c = -4

\(b^{2} − 4ac =3^{2} − 4(1)(−4) = 9 + 16 = 25\)

The quadratic formula loses quite a bit of its glory. \(x = \dfrac{-3 \pm \sqrt{25}}{2(1)} = \dfrac{-3 \pm 5}{2}\)

It's a WHOLE LOT EASIER all of a sudden, isn't it?!

Answer 36

yes

Answer 37

This one has an interesting implication. Without the complication of a square root in the final answer, you probably should feel a little bad about not factoring it. Given a little more effort, factoring can be found.

Okay, we're down to the last conversation. What if \(b^{2} - 4ac = 0\)??

What can we tell from that?

Answer 38

idk

Answer 39

Come on. I'm going to make you tell me. Look at the Quadratic formula and imaging what happens if you just throw out the square root part. \(\sqrt{0} = 0\), right?

Answer 40

ummmm......................................

Answer 41

Think about the numerator

-b + 0

-b - 0

What is so special about that?

Answer 42

b is negative

Answer 43

?? That makes no sense. Look at the three examples I provided above. b = 3.

What is so special about the relationship between

-3 + 0
and
-3 - 0

??

Answer 44

it still gonna be a ngative with both equations

Answer 45

Okay, if b = -5

What is special about the relationship between these two expressions?

-(-5) + 0 = 5+0
and
-(-5) - 0 = 5-0

Look at what changes in the two expressions. What changes and what does it do? Search your mind, Drag some sense out of the relationship!!

Answer 46

the addition and subtraction symbol

Answer 47

Okay, that's what changes. Does it make any difference if we are adding or subtracting zero (0)?

Answer 48

yes it changes it from being ngative/positive

Answer 49

It changes WHAT from being positive or negative?

-5 + 0 = ??
-5 - 0 = ??

Answer 50

the answer

Answer 51

Answer those last two questions. Fill in the two "??".

Answer 52

-5+0=-5
-5-0=5

Answer 53

Rethink!!! -5-0 = -5

How did you get +5?

If you can't get the arithmetic, you can't do the algebra. Please be more careful.

Answer 54

Note: You have been working very hard. You have been very patient. Hang in there just a little longer. You'll get it!

Answer 55

ok

Answer 56

Do you see that, now?

-5 + 0 = -5
-5 - 0 = -5

The are EXACTLY THE SAME!

Answer 57

Zero doesn't do anything in addition or subtraction.

Answer 58

yes i see and ok

Answer 59

im sorry i gtg will u b on 2morrow

Answer 60

Okay, this is a problem about the "discriminant". The "discriminant" is that thing that gives us information without having to solve the whole problem. In this case, the "discriminant" is just that \(b^{2} - 4ac\) that we have been beating to death for the last couple hours.

Recap of the discriminant
Negative? No Real Number Solution
Positive? Real Number Solutions
Perfect Square? Might have been able to factor it if we had tried harder.
Zero? Think to yourself, "I am a pansy, I failed to recognize a perfect square trinomial"

Whew!! You probably know more about the discriminant and the quadratic formula than anyone else in your class or study cohort.

Are we ready to solve the original problem? Do we remember what it was?

Find the value of the discriminant and describe the number and type of roots.
3x2 - 8x + 2 = 0

Go! Let's see what you get.

Answer 61

I'll be waiting for your answer.

Answer 62

@nincompoop

Answer 63

@mathslover

Answer 64

Jazzy, @tkhunny explained well . Where are you having problem?

Answer 65

i dont understand how to do it

Answer 66

@tkhunny idk i cant remember