Question

Quadratic Functions and Equations Unit

Answer 1

whats the question

Answer 2

@infinityknowledge

Answer 3

For the first one, use the quadratic formula. Do you know what that is?

Answer 4

yes

Answer 5

Than go ahead, plug in the numbers.

Answer 6

hint:
in order to solve thie equation: \(ax^2+bx+c=0\), we can apply the standard formula:
\[\Large x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]

Answer 7

i know the formula but idk how to plug the numbers in

Answer 8

Just subtract each of the variables into the formula.

Answer 9

for example, for question 1, we have:
\(a=2,\;b=1,\;c=2\)

Answer 10

whereas for question 2, we have:
\(a=1,\;b=2,\;c=-8\)

Answer 11

im sorry i just dont geti t

Answer 12

so like

Answer 13

please you have to substitute those values of a, b, c, into the standard formula, so, for first question we can write:

\[\Large \begin{gathered}
x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} = \hfill \\
\hfill \\
= \frac{{ - 1 \pm \sqrt {{1^2} - \left( {4 \cdot 2 \cdot 2} \right)} }}{{2 \cdot 1}} = ...? \hfill \\
\end{gathered} \]
please continue

Answer 14

@Michele_Laino 1+16/2

Answer 15

hint:
here is the next step:

\[\Large \begin{gathered}
x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} = \hfill \\
\hfill \\
= \frac{{ - 1 \pm \sqrt {{1^2} - \left( {4 \cdot 2 \cdot 2} \right)} }}{{2 \cdot 1}} = \hfill \\
\hfill \\
= \frac{{ - 1 \pm \sqrt {1 - 16} }}{2} = \begin{array}{*{20}{c}}
{\frac{{ - 1 + \sqrt {1 - 16} }}{2} = ...?} \\
{\frac{{ - 1 - \sqrt {1 - 16} }}{2} = ...?}
\end{array} \hfill \\
\end{gathered} \]

Answer 16

Ahhhhh
I got my signs wrong

Answer 17

It would be 15 then right?

Answer 18

\(1-16=-15\) am I right?

Answer 19

I think so

Answer 20

so, what are the solutions?

Answer 21

All the 15s are positive though

Answer 22

I think its b

Answer 23

sorry I have made a typo, here are the right steps:
\[\Large \begin{gathered}
x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} = \hfill \\
\hfill \\
= \frac{{ - 1 \pm \sqrt {{1^2} - \left( {4 \cdot 2 \cdot 2} \right)} }}{{2 \cdot 2}} = \hfill \\
\hfill \\
= \frac{{ - 1 \pm \sqrt {1 - 16} }}{4} = \begin{array}{*{20}{c}}
{\frac{{ - 1 + \sqrt {1 - 16} }}{4} = ...?} \\
{\frac{{ - 1 - \sqrt {1 - 16} }}{4} = ...?}
\end{array} \hfill \\
\end{gathered} \]

Answer 24

Okay, well would it still be -15?

Answer 25

yes! more precisely, we have to compute this:
\[\huge \sqrt { - 15} = ...?\]

Answer 26

okay whats next

Answer 27

in order to do that, it is necessary to introduce imaginary numbers

Answer 28

Ah okay

Answer 29

so we have:
\[\large \sqrt { - 15} = \sqrt {\left( { - 1} \right) \cdot 15} = \sqrt { - 1} \cdot \sqrt {15} = ...?\]

Answer 30

i15?

Answer 31

better is:
\[\Large \sqrt { - 15} = \sqrt {\left( { - 1} \right) \cdot 15} = \sqrt { - 1} \cdot \sqrt {15} = i\sqrt {15} \]
since, by fdefinition, we have:
\[\huge \sqrt { - 1} = i\]

Answer 32

definition*

Answer 33

so, what is the right option?

Answer 34

c! :)

Answer 35

that's right!

Answer 36

Thank you so much

Answer 37

:)
now, please apply the same procedure for second equation

Answer 38

its
b

Answer 39

please wait, I'm checking your computation

Answer 40

I think that option b is wrong

Answer 41

wait wait

Answer 42

Its a, sorry/

Answer 43

that's right!

Answer 44

Wooo hoooo!

Answer 45

Can you help me with the other one?

Answer 46

ok!

Answer 47

Thank you!

Answer 48

as we can see we have two zeroes:
\(x_1=-2\) and \(x_2=-1\)
we can write this:
\[\begin{gathered}
s = {x_1} + {x_2} = - 2 - 1 = - 3 \hfill \\
p = {x_1} \cdot {x_2} = \left( { - 2} \right) \cdot \left( { - 1} \right) = 2 \hfill \\
\end{gathered} \]
now you have to search for an equation like this:
\[\huge {x^2} - sx + p = 0\]

Answer 49

search for one?

Answer 50

yes! Please substitute the value of \(s\) and \(p\) first

Answer 51

hint:
you have to replace \(s=-3\) and \(p=2\) into this equation:
\[\huge {x^2} - sx + p = 0\]

Answer 52

Uhhhhh..you lost me there

Answer 53

what do i do to sub

Answer 54

hint:
pleas consider this equation:
\[\huge {x^2} - sx + p = 0\]
then rewrite it, when \(\Large s=-3\) and \( \Large p=2\)

Answer 55

please*

Answer 56

x^2 - 1?

Answer 57

another step:
after I replace \(\Large s=-3\), I get:
\[\huge {x^2} - \left( { - 3} \right)x + p = 0\]
now, replace \(\Large p=2\)

Answer 58

Yes and then i add them right?

Answer 59

Thats what i did

Answer 60

please what equation do you get?

Answer 61

x^2 - (2)1 = 0

Answer 62

I got this:
\[\huge {x^2} + 3x + p = 0\]
now, please replace \(\Large p=2\), what do you get?

Answer 63

Yes. x2+3x+2=0

Answer 64

ok! So, what is the right option?

Answer 65

no solution?

Answer 66

are you sure?

Answer 67

no but since the equation equals 0

Answer 68

yes! so what option does contain the equation you got

Answer 69

cannot be determined?

Answer 70

is option A correct?

Answer 71

im not sure. you questioned me lol

Answer 72

please look at your question:

Answer 73

geez, where working on two different questions!

Answer 74

im sorry

Answer 75

no problem :)

Answer 76

so is it A?

Answer 77

no, since option A contains the equation \(-x^2+3x+2=0\)

Answer 78

our equation is: \(x^2+3x+2=0\)

Answer 79

c since its on negative 2?

Answer 80

option C is a wrong option

Answer 81

i dont understand then

Answer 82

is option B correct?

Answer 83

Yes? that the equation we came up with right?

Answer 84

that's right!

Answer 85

Thank you ao much again

Answer 86

:)