Question

Solve the following system of equations and show all work.

y = −x2 + 4
y = 2x + 1

Answer 1

I can walk you through this, ok?

Answer 2

ok

Answer 3

Firstly, do you think we could set these equal?

Answer 4

Ie Y = Y?

Answer 5

whats le?

Answer 6

ie, in other words

Answer 7

ok

Answer 8

y = −x2 + 4
y = 2x + 1

Both Y, set them equal to each other =)

Answer 9

-x2 + 4 = 2x +1

Answer 10

Correct.

Now, any idea what we could do?

Answer 11

nope

Answer 12

Ever heard of a quadratic equation?

Answer 13

yes, but i don't remember much

Answer 14

A quadratic takes the form ax2 + bx + c = 0

Answer 15

Can you see how we could manipulate -x2 + 4 = 2x +1

To reach ax2 + bx + c = 0 ?

Answer 16

-x2 + 2x + 5= 0

Answer 17

Not quite. We can't have a negative x squared.

Answer 18

oh okay

Answer 19

So if we moved it across the equals?

Answer 20

idk

Answer 21

what would we move?

Answer 22

Lets get x squared positive, and manipulate to get a ax2 + bx + c = 0 form

Answer 23

okay

Answer 24

What do you get?

Answer 25

Am i changing the whole equation?

Answer 26

-x2 + 4 = 2x +1 => ax2 + bx + c = 0 form

Answer 27

2x-x2+5

Answer 28

Just a moment

Answer 29

ok

Answer 30

Are you still here? :)

Answer 31

Yes, was just about to ask you the same question

Answer 32

Ok, lets show you how this part is done.

Answer 33

ok

Answer 34

-x2 + 4 = 2x +1

1.) Our x squared is negative, so lets make it positive. We do this by bringing it across the equals.

= 4 = x^2 + 2x + 1

Cool?

Answer 35

okay

Answer 36

2.) Now we deal with the X.

2x is the only x in the equation so it's fine. It needs to be beside x^2 to satisfy our ax2 + bx + c = 0 form, which it is.

4 = x^2 + 2x + 1

Answer 37

So we basically just moved it back?

Answer 38

It didn't move at all. We just brought x^2 over.

I'm just showing how we didnt need to touch it.

Answer 39

Oh i see it

Answer 40

You just took away the "=" sign at the beginning

Answer 41

Are you with me so far?

Answer 42

i think so

Answer 43

A recap.

1.) We have two equations both Y =

For this reason, we can set them equal to each other.

2.) We then can see that we have x^2, 2x and constants, for this reason, we can see that we can manipulate ''-x2 + 4 = 2x +1'' into a quadratic in the form ax^2 + bx + c = 0.

3.) We deal with the X^2 first, then X, then constants.

4.) The X^2 cannot be negative, we move it over the equals.

5.) The 2x is fine, and satisfies the form ax^2 + bx + c = 0 as it is.

Ok?

Answer 44

ok, does it matter whether we picked the "+4" or the "+1"?

Answer 45

We haven't gotten to the constants just yet.

Answer 46

4 = x^2 + 2x + 1

This is where we are.

Our X^2 is good!
Our 2x is good!

Now we have a 4 on the left and a 1 on the right.

Answer 47

i mean, if we did 1 = x^2 + 2x + 4, would that matter?

Answer 48

In essence, the final solution will be x^2 + 2x -3 = 0

Answer 49

y = −x2 + 4
y = 2x + 1

make y = y and will get

-x^2 +4 = 2x +1 divide both sides by -1 and will get

x^2 -4 = -2x -1 add to both sides 2x+1 and will get

x^2 +2x +1 -4 = 0

x^2 +2x -3 = 0 how you like solve it ? using discriminant or by factored ?

Answer 50

Embrace, can you see how we get x^2 + 2x - 3 = 0?

Answer 51

nice job - good luck

Answer 52

hold on

Answer 53

yeah i see it

Answer 54

So, this is our final solution.

It is a quadratic equation.

Answer 55

Now, do you know how to solve a quadratic?

Answer 56

we graph it?

Answer 57

In this case, we want to factorise it, to get two solutions.

Answer 58

But you're not wrong either, we COULD graph it.

Answer 59

okay

Answer 60

I'll go with factoring

Answer 61

It would look something like this.

Answer 62

If we graphed :)

Answer 63

ok, i still prefer factoring beacuse i have to explain this in words and i dont wanna have to explain a graph

Answer 64

You should always factorise unless asked to graph. :)

Answer 65

ok

Answer 66

We want two numbers that multiply to -3 and add to 2.

Answer 67

3 and -1

Answer 68

Correct!

If we were to split up the middle just to show:

x² - x + 3x - 3 = 0

With me?

Answer 69

kind of

Answer 70

Have you ever factorised a quadratic before? :)

Answer 71

yes

Answer 72

Ok.

So you said 3 and -1.

This would mean our

x^2 + 2x -3 = 0

becomes:

x² - x + 3x - 3 = 0

Answer 73

The -1 turned into an -x?

Answer 74

They are equivalent.

-1 = -x in this case.

Answer 75

You said 3 and -1

-x is the equivalent of your -1.
3x is the equivalent of your 3.

:)

Answer 76

I am lost XD

Answer 77

\[x^2 + 2x -3 = 0\]

Answer 78

I am so sorry, i dont know what step in solving this we're on XD

Answer 79

It's no problem.

Lets just go back to our x^2 + 2x -3 = 0

Ok? :)

Answer 80

okay

Answer 81

In quadratic equations which are trinomial, the following is true:

> The product of the roots is the constant term.
> The sum of the roots is the coefficient of the middle term.

Answer 82

Eg.

x^2 - 12x + 27 = 0

What would our two roots be?

Answer 83

We want 2 numbers that multiply to give +27, and add/subtract to give -12.

Answer 84

ok