Question

i will become a fan and i will give you a medal if you can help me with this question

Answer 1

Felix exclaims that his quadratic with a discriminant of −1 has no real solutions. Felix then puts down his pencil and refuses to do any more work. Create an equation with a negative discriminant. Then explain to Felix, in calm and complete sentences, how to find the solutions, even though they are not real.

Answer 2

i dont know this awnser

Answer 3

Yeah i don't know this either

Answer 4

@Preetha @whpalmer4 @Zedd @Dobby1 @shamil98

Answer 5

Do you know what the discriminant is?

Answer 6

its a formula right

Answer 7

If you have a quadratic:
\[ax^2+bx+c =0\]The discriminant is \(b^2-4ac\), which you may recognize as the thing under the radical sign in the formula for the solutions of a quadratic:
\[x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Answer 8

yes

Answer 9

If we have a "perfect square", such as \((x+2)^2 = x^2+4x+4\), then the discriminant is equal to 0, and our solution is \[x = -\frac{b}{2a}\]You might even recognize that as the formula for the x-coordinate of the vertex of a parabola.

Answer 10

If we don't have a perfect square, and the discriminant is positive, then our solutions look like\[x = \frac{-b\pm d}{2a}\]where \(d\) is a positive number.

Answer 11

nope not really

Answer 12

If we graph the equation, those values of \(x\) will be the points at which the parabola crosses the x-axis, also known as the zeros or the roots or the solutions.

Answer 13

And the final case is where the discriminant is negative. What happens when you take the square root of a negative number?

Answer 14

ok well the discriminate number is negative 1 and it dont have a perfect square

Answer 15

so like what do we do

Answer 16

I asked you a question: what happens when you take the square root of a negative number?

Answer 17

idk

Answer 18

how on looking through my notes now

Answer 19

yea i cant seem to find it

Answer 20

Well, let's do some thinking. What is the square root of 4?

Answer 21

2

Answer 22

Okay, and we know that is true because 2*2 = 4.

What is the square root of -4?

Answer 23

-2

Answer 24

I thought you might say that. What does\[-2*-2 =\]

Answer 25

oh it would be a postive 4

Answer 26

Right! What kind of a number can we multiply by itself and get a negative number?

A positive number * a positive number gives us a positive number.
A negative number * a negative number also gives us a positive number.

Answer 27

a negative and a positive

Answer 28

Ah, but then it won't be multiplying by itself, will it?

Answer 29

oh yea...i can think of anything

Answer 30

*cant

Answer 31

It's like we need to have some sort of number we can multiply by itself and get -1. Then we could factor the square root of -4 like this:
\[\sqrt{-4} = \sqrt{-1*4} = \sqrt{-1}*\sqrt{4} = 2\sqrt{-1}\]We could give this special number a name, \(i\), and use it like a variable when we do algebra, but always remembering that \[i=\sqrt{-1}\]\[i^2=-1\]

Answer 32

\[\sqrt{-4} = \sqrt{4}*\sqrt{-1} = 2i\]

Going the other way:
\[2i*2i = 4i^2\]but by definition, \(i^2=-1\) so that becomes
\[2i*2i = 4i^2 = 4*-1 = -4\]And that's exactly what we want from the square root of -4, right?

Answer 33

right

Answer 34

Going back to our problem, if we have a negative discriminant \((b^2-4ac < 0)\), that means we'll have a square root of a negative number in our solutions to the quadratic equation. A square root of a negative number is an imaginary number (which is why we chose \(i\)). An imaginary number combined with a real number is a complex number, which can be written in the form \(a+bi\) where \(a,b\) are real numbers and \(i = \sqrt{-1}\)

Answer 35

Shall we do an example of solving a quadratic that has imaginary or complex solutions?

Answer 36

yes please

Answer 37

Let's pick one where \[a=1,\,b=2,\,c=3\]

\[y=1x^2+2x+3\]\[y=x^2+2x+3\]

What will the value of the discriminant be?

Answer 38

now wouldnt we use the quadratic formula

Answer 39

please, when I ask you a question, do me the courtesy of answering it...

Answer 40

\[\sqrt[-2]{-8} \over 4\]

Answer 41

No, I'm not sure what that is supposed to be, but it isn't the discriminant :-)

If your quadratic is in the form \[ax^2+bx+c=0\]the discriminant is \[\Delta =b^2-4ac\](that's a Greek letter called delta, and it is often used as a symbol for the discriminant)

What is the value of the discriminant for \[x^2+2x+3=0\]\[\Delta=\]?

Answer 42

it would be -8

Answer 43

very good!

Answer 44

yayyyy

Answer 45

So let's proceed on to find the solutions. Can you fill in the blanks in the quadratic formula?

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

Answer 46

actually, before you do that, let's just focus on the \(\sqrt{\Delta}\) part.

You found that \(\Delta=-8\), so we need to find the value of \[\sqrt{-8}\]Can you simplify that at all?

Answer 47

no

Answer 48

and i had already put in the quadratic equation

Answer 49

Can you simplify \[\sqrt{8}\]?

Answer 50

4

Answer 51

Does 4*4 = 8?

Answer 52

1*8=8

Answer 53

What's your point?

Answer 54

Are those the only numbers which you can multiply to get 8?

Answer 55

2*4

Answer 56

Right. Can you take the square root of either of those numbers?

Answer 57

\[\sqrt{8} = \sqrt{2*4} = \sqrt{2}*\sqrt{4} = \]

Answer 58

4 which is 2

Answer 59

So that simplifies to...

Answer 60

2

Answer 61

No, \[\sqrt{8} \ne 2\]\[\sqrt{8} = \sqrt{2}*\sqrt{4} = \]

Answer 62

\[\sqrt{8}=\sqrt{2}*\sqrt{4}=\sqrt{2}*2=2\sqrt{2}\]

Answer 63

But we have \[\sqrt{-8}\]How do we form the square root of a negative number?

Answer 64

Remember that example I did, where I showed that \[\sqrt{-4} = 2i\]?
What does that suggest to you?

Answer 65

oh yea i had forgot about that

Answer 66

okay, well, reread that section of the book we're writing here and tell me what the answer is...

Answer 67

i dont have a book

Answer 68

"book we're writing here" — scroll back and read the example I mentioned...

Answer 69

\[\sqrt{-8}=4i\]

Answer 70

:-(

If that's true, then \[4i*4i=-8\]\[4i*4i=4*4*i*i = 16i^2 = 16(-1) = -16\]

Answer 71

ughhhhhh

Answer 72

Look, \[\sqrt{-a} = \sqrt{-1*a} = \sqrt{-1}*\sqrt{a} = i\sqrt{a}\]So if \[\sqrt{8} = 2\sqrt{2}, \sqrt{-8} =\]

Answer 73

-2 sqrt 2

Answer 74

I think you should go to this web site: https://www.khanacademy.org/math/arithmetic/exponents-radicals

watch the videos in the section labeled "The Square Root"

Answer 75

No, that's not correct.

\[-2\sqrt{2}*-2\sqrt{2} = -2*-2*\sqrt{2}*\sqrt{2} = 4*2 = 8\]

Answer 76

Again, if the square root of a positive number is \(a\), then the square root of the negative number will be \(ai\).

square root of 8 is \(2\sqrt{2}\)
square root of -8 is \(2\sqrt{2}i \text{ or } 2i\sqrt{2}\)

Answer 77

This might also help: https://www.khanacademy.org/math/trigonometry/imaginary_complex_precalc/i_precalc/v/imaginary-roots-of-negative-numbers