Question

Medal + Fan For helping solve this.

[9.05] Part 1: Create your own quadratic equation that cannot be solved by factoring, but can be solved using the quadratic formula. Identify the values of a, b, and c, and find the solutions using the quadratic formula. Show all work to receive credit.

Part 2: Using complete sentences, explain how you know that the equation from Part 1 cannot be solved by factoring, but can be solved by using the quadratic formula.

Answer 1

@amistre64

Answer 2

@amistre64 Please! I will Be fan + Medal This is my last question out of 25 I cannot figure it out!

Answer 3

you might want to start by showing me what you think the quadratic formula looks like

Answer 4

there is one part of the formula that is critical to the question, namely:

b^2 - 4ac

if this setp is negative, there are no "Real" solutions

Answer 5

*setup

Answer 6

I don't even know what a quadrant equation is.
I really would like to actually learn this stuff. But it's hard!

Answer 7

How is it that you are even doing this exercise without knowing what the quadratic equation is?

Answer 8

I tried to read my lesson, but I did not get it.

Answer 9

Like I said, I DO want to learn this. I don't just want the answer.

Answer 10

You first need to understand that what you are doing is creating an equation that takes the form \[ax^{2}+bx+c=0\]

Answer 11

You are familiar with factoring?

Answer 12

Yes I am familiar with factoring.

Answer 13

Okay, give me a quadratic that you can factor.

Answer 14

x2 − 3x = 28
Is that a quadratic?

Answer 15

Almost, you have to be equal to 0 so it would need to be \[x^2-3x-28=0\]

Answer 16

Okay. Next what?

Answer 17

The problem statement wants you to create one that CANNOT be solved by factoring, so what could you do to make this unable to factor?

Answer 18

It might help to think about what about it allows you to factor it.

Answer 19

ax^2 + bx + c = 0?

Answer 20

This is a form I think. Is it factorable?

Answer 21

ax^2 + bx + c = 0 is factorable on the Reals if and only if, b^2 -4ac is not a negative value

Answer 22

the quadratic formula is:\[x=\frac{-b\pm\color{red}{\sqrt{b^2-4ac}}}{2a}\]

Answer 23

as long as that under the radical is 0 or positive, we can factor it into real numbers

Answer 24

So using that formula how would we add real numbers in it?

Answer 25

well, first lets define our favorite values for a and c .... you got any 2 number that make you happy?

Answer 26

None really, any under 10 will do

Answer 27

this is yours to work out, you pick ....

Answer 28

How about 2 and 7

Answer 29

:)

Answer 30

Okay, my pick... ummmmm. 2 and 4

Answer 31

lol, great

b^2 - 4(2)(4) < 0

b^2 < 4(2)(4)

b^2 < 32

you know of any squares less than 32?

Answer 32

No.

Answer 33

you know of any squares greater than 32?

Answer 34

I don't know what you mean?

Answer 35

b^2 is notation for b squared

when we square a number, we multiply it to itself

0*0 = 0
1*1 = 1
2*2 = 4
3*3 = 9
4*4 = 16
5*5 = 25
6*6 = 36
....

any of these squares less than 32 will work

Answer 36

3*3

Answer 37

great :)

so, a = 2, c = 4, and b= 3

ax^2 + bx + c = 0
2x^2 + 3x + 4 = 0 ... is not factorable under the set of Real numbers

Answer 38

That makes a little more since!

Answer 39

good luck :)

im just a little curious what "factorable by the quad formula, but not by factoring" might mean.

Answer 40

It probably means they want one that you MUST use the quadratic formula for.

Answer 41

Hey, Thanks so much mate! Very grateful! To the both of you...
@eSpeX
@amistre64

Answer 42

nothing NEEDS the quadratic formula tho ... thats just the result of completing the square

Answer 43

You're welcome @nicksolis it's nice to find someone who appreciates learning.

Answer 44

True that you don't "need" it, but think of the context, you have people learning by factoring and are all of the sudden exposed to the quadratic formula.

Answer 45

does it mean "integer" factors by chance? or at best rational factors?

Answer 46

I am not sure.

Answer 47

if so, then we could simply assert that sqrt(b^2-4ac) not be a perfect square

Answer 48

\[\frac{-b\pm\sqrt{k^2}}{2a}=\frac{-b-k}{2a},\frac{-b+k}{2a}\]
whereas
\[\frac{-b\pm\sqrt{p}}{2a}=\frac{-b-\sqrt{p}}{2a},\frac{-b+\sqrt{p}}{2a}~:~p\ne Q\]

Answer 49

That would be inline with my thinking. Just something a student could not easily do in their head.

Answer 50

Anyway, good luck @nicksolis