- jamiebookeater

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- pooja195

- mathmate

Finally, got here!

- mathmate

Start with quadratic equations, ok?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- pooja195

ok :)

- mathmate

Solve
\(x^2 -6x +9=0\)

- pooja195

a=1
b=-6
c=9

- mathmate

ok.

- pooja195

\[\huge~x=\frac{ -(-6)\pm \sqrt{(-6)^2-4(1)(9)} }{ 2(1) } \]

- mathmate

Great, now can you simplify that, because it will be greatly simplified when you put the numbers in there.

- pooja195

\[\huge~x=\frac{ 6\pm \sqrt{0} }{ 2}\]

- pooja195

\[\huge~\frac{ 6 }{ 2 }=3~~~~~x=3\]

- mathmate

Yes, that is the same as {3,3}, or 3 (twice)

- mathmate

It's called a double root, andthe graph looks like this:|dw:1433211976391:dw|
the vertex just touches the x-axis.

- mathmate

ok so far?

- pooja195

ye

- mathmate

If you factored, you would have got (x-3)^2=0, which is the same as saying (x-3)=0, or
x=3 (twice), or x={3,3}

- mathmate

Was wondering you still have the online textbook?

- mathmate

Can you dig some questions out of there, which will be more like what you have learned.

- pooja195

T_T but there are answers ,_,

- mathmate

You don't have to look at them, in any case, we are looking at the work, can check the answers at the end.

- mathmate

There may be word problems, which take more time to make. Using ready-made problems can help us do more.

- pooja195

##### 1 Attachment

- mathmate

Have you done anything like this?
"A picture has a height that is 4/3 its width. It is to be enlarged to have an area of 192 square inches. What will be the dimensions of the enlargement?"

- pooja195

:/

- pooja195

no O_O

- mathmate

Always the numerical equations, and nothing like word problems?

- pooja195

no ._.

- mathmate

ok, so we can skip that.

- pooja195

:)

- mathmate

Whenever you solve quadratics, there are three ways to do it.
If the teacher does not specify, you can use any method you want. If he/she specifies completing the square, you have no choice.
the three are:

- mathmate

quad formula (just did)
factoring (did over the weekend)
completing the square
Do youknow how to do the last one?

- pooja195

no :/ we havent learned it

- mathmate

ok, so we skip that too.
Anything else you'd like to do on quadratic equations/

- pooja195

Nope.

- mathmate

So we will move onto system of linear equations.
Have you done word problems?

- pooja195

yes but dont like those :/

- mathmate

How many ways do you know to solve system of linear equations?

- pooja195

Elmination
Substituion
:/

- mathmate

comparison?

- pooja195

eh?

- mathmate

When you have
y=2x+3
y=4x-2
then you equate the right-hand sides (since they both equal y) to give
4x-2=2x+3 and solve for x.
It is a form of elimination, but makes life easier. (called method of comparison)

- pooja195

i know that :P

- mathmate

Ok, so let go:
Solve
y = 36 â€“ 9x
3x + y/3 = 12

- mathmate

*let's

- pooja195

3x+(36-9x)/3=12
6x+36/3=12
6x+12=12
x=0

- mathmate

...not finished! need y, and the final answer!

- pooja195

y=36-9(0)
y=36-0
y=36?
(0,36)

- mathmate

Excellent!

- mathmate

Solve the following system by substitution.
2x â€“ 3y = â€“2
4x + y = 24
When you see a singleton y, or singleton x, then you can try substitution, as in the above case.

- pooja195

y=-4x+24
2x-3y=-2
2x-3(-4x+24)=-2
2x+12x-72=-2
14x-72=-2
14x=70
x=5
y=-4(5)+24
y=20+24
y=44
Final answer
(5,44)

- mathmate

Something wrong here, otherwise all correct.
y=-4(5)+24
y=20+24
y=44

- pooja195

y=4

- mathmate

yep!
Solve the following system using addition (a kind of elimination)
2x + y = 9
3x â€“ y = 16

- pooja195

x=5
2(5)+y=9
10+y=9
y=-1
(5,-1)

- mathmate

That was fast! You solved for x by inspection! But show work in exams!

- mathmate

The admission fee at a small fair is $1.50 for children and $4.00 for adults. On a certain day, 2200 people enter the fair and $5050 is collected. How many children and how many adults attended?

- pooja195

O_O

- pooja195

Nooooooooooooooooo

- mathmate

With word problems, always define variables first, so you don't get confused at the end.

- mathmate

I'll do this one, ok?

- mathmate

Define variables:
Number of children = c
Number of adults = a
(we'll could get confused using x,y, but never this way)

- mathmate

Admission for children = $1.50
Admission for adults = $4

- mathmate

"2200 people enter the fair and $5050 is collected"
means
c+a = 2200
1.50c + 4a=5050 = 15c+40a = 50500 (keep integers as long as possible
Substitute (because c has a coefficient of 1)
c=2200-a
so
15(2200-a) + 40a = 50500
expand
33000 -15a + 40a = 50500

- mathmate

40a-15a = 50500-33000
25a = 17500
a=17500/25 = 700
c=2200-700=1500
Check
1.5c+4a=2250+2800=5050 good!

- mathmate

You will find that once the variables are defined, everything else look easy.
What do you think?

- pooja195

It seems so complicated >_<

- pooja195

*-*

- mathmate

Do you want to tell me which part is harder to understand?

- mathmate

It looks complicated probably because I show all the work. If I skipped or jumped some steps, it will look easy, but a little harder to follow.

- pooja195

hmm.... :/

- mathmate

Now your turn!
The sum of the digits of a two-digit number is 7. When the digits are reversed, the number is increased by 27. Find the number

- mathmate

First define variables, use meaningful names, like
t = tens digit
u = units digit.
so far so good?

- mathmate

When you have a number like 25, then t=2, u=5, and
10t+u = 25.

- mathmate

"The sum of the digits of a two-digit number is 7"
can you form an equation with that?

- pooja195

7u+t=27?

- mathmate

Here, we are saying the equivalent of 2+5=7, so
t+u=7 will do.
now
"When the digits are reversed, the number is increased by 27. "

- pooja195

t+u=27?

- mathmate

10t+u is the original number
10u+t is the number reversed (units digit becomes 10's digit)
so
10u+t = 10t+u + 27 (increased by 27)

- mathmate

Simplifying this equation, we get
9u-9t=27
or
u-t=3
combined with
u+t=7
we solve for t=2, u=5, or the original number is 25.

- mathmate

Check: 52-25=27, ok.

- pooja195

:/

- mathmate

- pooja195

.

- mathmate

I'm typing up an example.

- mathmate

\(\large \frac{5}{x^2-1}+\frac{4}{x^2+2x+1}\)

- mathmate

So first factorize the denominators:

- mathmate

\(\frac{5}{(x+1)(x-1)}+\frac{4}{(x+1)^2}\)

- mathmate

Can you find the common denominator?
between (x+1)(x-1) and (x+1)^2

- pooja195

it would be (x-1)

- pooja195

and then x+1 on the other side?

- mathmate

"common" denominator of two expressions must contain ALL the factors, and is the same for all the terms.
So start with one of the two , say (x+1)(x-1) and add factors to it so that it contains every expression.

- mathmate

Say we start with (x+1)(x-1), surely it contains (x+1)(x-1) , which is the first expression.

- mathmate

Question is, does it contain the second expression (x+1)^2?

- mathmate

the answer is no.
What do we need to do? Multiply by another (x+1), so we end up with a common denominator of (x+1)(x-1)(x+1).

- mathmate

So going back to the question:
\(\Large \frac{5\color{red}{(x+1)}}{(x+1)(x-1)\color{red}{(x+1)}}+\frac{4\color{blue}{(x-1)}}{(x+1)^2\color{blue}{(x-1)}}\)
notice that now both denominators are identical.

- mathmate

\(\Large \frac{5\color{red}{(x+1)}}{(x+1)^2(x-1)}+\frac{4\color{blue}{(x-1)}}{(x+1)^2(x-1)}\)

- mathmate

and that's the same as
\(\Large \frac{5\color{red}{(x+1)}+4\color{blue}{(x-1)}}{(x+1)^2(x-1)}\)

- mathmate

Next step is to expand and simplify the numerator:
\(\Large \frac{5x+5+4x-4}{(x+1)^2(x-1)}=\frac{9x+1}{(x+1)^2(x-1)}\)
and that's the end of the calculations.

- pooja195

I read these but only half .-.
@mathmate

- mathmate

Did you understand as far as you read?

- pooja195

yes but now we can start studying for the finals :P >_<

- mathmate

Here's the diagnostic test, as far as I have done.
I will add other things on.

##### 1 Attachment

- mathmate

@pooja195 you there?

- pooja195

i iz here

- mathmate

ok, let's get started.
Any questions on chapter 1?

- mathmate

Say, 4^d=64, what is "d".

- mathmate

need help?

- pooja195

yeah :/

- mathmate

4^0=?

- pooja195

1

- mathmate

Very good!
4^1=

- pooja195

4

- mathmate

Just two rules:
Anything raised to the power of 0 is 1,
anything raised to the power of 1 is the base itself, like 4^1=4

- mathmate

4^2=

- pooja195

16

- mathmate

Remember,
\(4^2=4.4\),
\(4^3=4.4.4\),
\(4^4=4.4.4.4\), etc.

- mathmate

so 4^3=?

- pooja195

64

- pooja195

so 3?

- mathmate

compare
4^3=64, 4^d=64
so what is d? yes, d=3

- mathmate

Great!
Now try 8=2^z. what is z?

- pooja195

2*2*2
2^3

- mathmate

and z=?

- pooja195

3

- mathmate

Good!
Now a difficult one: \(-3^2=\)?

- pooja195

-9?
or if its (-3)^2 it would be 9

- mathmate

Exactly, that's where many students make mistakes.
when in doubt, use PEMDAS which says that exponentiation goes before subtraction (negative), so what you did was correct.
(-3)^2 would raise the power of -3, so (-3)*(-3)=+9.
Well done!

- mathmate

and -3^2=-9.

- mathmate

Try the following:
1^3=
3=3^j, j=
1=4^h, h=
3^r=81, r=
(-2)^3=

- mathmate

Laws of exponent:
2^n = 2.2.2.....2 (n times)
\(a^m \times a^n = a^{m+n}\)
\(\large a^{-m}=\frac{1}{a^m}\)
\(\large a^{\frac{1}{2}}=\sqrt a\)
\(\Large a^{-\frac{1}{2}}=\frac{1}{\sqrt a}\)

- mathmate

Now practice:
\(\Large (\frac{1}{2})^2=\)
\(\Large (0.2)^3=(\frac{2}{10})^3=(\frac{1}{5})^3=\frac{1}{125}\)
Express \(\large 3^{-8}\) with a positive exponent. \(\large 3^{-8}\) =

- pooja195

O_O

- mathmate

|dw:1433336557931:dw|

- pooja195

What is that??

- mathmate

Helps you see where numbers belong.
R includes all numbers you're studying (i.e. does not include complex numbers)
Inside of R is divided into Q (rational) and Q' (irrationals)
Q' (irrationals) include numbers like sqrt(2), \(\pi\), sin(10\(^\circ \)), etc.
Q (rationals) include \(all\) integers (Z), and all naturals (N).
Decimals (2.5) repeating decimals (2.33333...), fractions (7/3) all belong to rationals, but not Z and N.
All N belong to Z, but negative integers are in Z but not in N.
See if you can read these relations from the diagram.

- mathmate

- pooja195

:)

- mathmate

Solve x+4<7 (page 324)

- pooja195

x<3

- mathmate

ok... too easy for you?

- pooja195

Yes xD

- mathmate

Write an inequality to show the max. no. of apples I can buy if I have $4.8 and apples cost $0.60 each.

- mathmate

* possible number of apples

- pooja195

O_O i dunno this...

- pooja195

x+0.60<4.8?

- mathmate

I have $4.80, and apples cost 0.60 each.
So the maximum number of apples I can buy is $4.80/0.60 =?

- pooja195

8

- mathmate

Good, and the minimum?

- pooja195

would i subtract?

- mathmate

Well, the minimum would be zero. I don't have to buy any.
So the answer would be:

- pooja195

x=8?

- pooja195

x<8

- mathmate

That's the maximum number of apples.
If we ask for the possible number, then it would be:

- pooja195

4? o.L

- mathmate

I can buy anything from 0 to 8, so
\(\large 0\le x \le 8\), where x is the number of apples.

- pooja195

ooo >_<

- mathmate

Does that make sense?

- pooja195

Yea

- mathmate

Word problems!

- mathmate

Try 4x-3=21

- mathmate

solve.

- pooja195

x=6

- mathmate

Excellent! now solve
-5x+10=30

- pooja195

x=-4

- mathmate

Good!

- mathmate

now solve
\(-5x+10\le30\)

- pooja195

\[\huge~x \ge-4\]

- mathmate

Very good!!! you actually know the rules!

- mathmate

Why is it \(\ge\) and not \(\le\) ?

- pooja195

because when you divide by a negative you always need to flip the signs

- mathmate

Excellent! What if you multiply by a negative?

- pooja195

Keep the sign as it is DO NOT change it :)

- mathmate

Is x/(-1) equal to x*(-1)?
why would the rule be different?

- pooja195

idk its just like that .-.

- mathmate

Actually, whether you multiply or divide by a negative, you flip the direction of the > or < sign. You can remember that precisely because x*(-1) is the same as x/(-1).
So can you now solve

- mathmate

\(\Large \frac{5}{3}(x-5)=6\)

- pooja195

\[\frac{ 5 }{ 3 }x-\frac{ 25 }{ 3 }=6\]

- mathmate

There is a trick to solving equations with fractional coefficients!

- mathmate

I would multiply by the denominator, or LCM of the denominator if there are many.

- mathmate

This way, we end up with an equation with integer coefficients.

- mathmate

\(\Large \frac{5}{3}(x-5)=6\)
\(\Large 3*\frac{5}{3}(x-5)=3*6\)
\(\Large 5(x-5)=18\)
and then go from here

- pooja195

\[\huge~5x-25=18\]
\[\huge~5x=43\]
\[\huge~x=\frac{ 43}{ 5 }\]

- mathmate

Exactly!

- mathmate

Any questions before we move on to linear equations?

- mathmate

...and graphing?

- pooja195

no, but lets skip graphing :/

- mathmate

You're willing to forfeit the points?

- pooja195

nvm -_-

- mathmate

What is a linear equation in slope-intercept form?

- pooja195

y=mx+b

- mathmate

Good! Can you find a line with a slope of 2 that passes through (2,4)?

- pooja195

Point slope formula
\[\huge~y-y1=m(x-x1)\]
\[\huge~y-4=2(x-2)\]
\[\huge~y-4=2x-4\]
\[\huge~y=2x\]

- mathmate

Excellent.
What form of equation did you use?

- mathmate

I call it point-slope form.
Not many school teach this form.

- mathmate

*schools

- pooja195

i wrote it on top xD

- mathmate

Sorry, I didn't see it!

- pooja195

itz okz :-)

- mathmate

Now how about a line that passes through 2 points. Can you find it?

- mathmate

The points are (2,5), (7,15)

- pooja195

Find slope first
\[\huge~\frac{ y2-y1 }{ x2-x1 }\]
\[\huge~\frac{ 15-5 }{ 7-2}=\frac{ 10 }{ 5 }=5\]
Point slope formula
\[\huge~y-y1=m(x-x1)\]
Pick a point right?
\[\huge~y-5=5(x-2)\]
\[\huge~y-5=5x-10\]
\[\huge~y=5x-5\]

- mathmate

you got 90% for this one!

- pooja195

-_-

- mathmate

Spot the error!

- pooja195

i dont see it .-.

- pooja195

Wait i simplified

- mathmate

All the steps are perfect, but 10/5=2 !!!!

- pooja195

>_<

- pooja195

Are you going to make me redo all the work????? ;-;

- mathmate

That's no problem. That's how I would correct, more for steps.

- mathmate

No!

- pooja195

:)

- mathmate

If your steps are good, that's what's important.

- mathmate

But do be careful in tests and exams.

- pooja195

ok :)

- mathmate

Translate into a mathematical inequality:
Ken's age (k) is at least four years older than Pete's (p).

- pooja195

\[\huge~K \ge 4+p\]

- pooja195

:/

- mathmate

Great! It wasn't easy for you, but you did it!

- mathmate

Now try:
Liz (L) has at least 4 times less flowers than Priya (P).

- mathmate

At least 4 times could mean 5 times!

- mathmate

* could be

- pooja195

Idk know this one .-.

- mathmate

When in doubt, put numbers in.
Say Liz has $10, how much should Priya have?

- mathmate

$40, $50, $60, anything $40 or more, right?

- mathmate

Sorry, change flowers to $...:(

- mathmate

So if L=10, P=40, so we write
\(\Large 4L \le P\)

- pooja195

i knew it!

- mathmate

That's what you would have done?

- pooja195

i was thinking something similiar

- pooja195

4p

- mathmate

Good. Be careful, many students would write L\(\le\) 4P.

- pooja195

XD

- mathmate

so using numbers, you can check your work.

- pooja195

Right

- mathmate

How are we doing so far?

- pooja195

Great :)

- mathmate

Pace? 1=too slow, 10=too fast.

- pooja195

5

- mathmate

ok, we're on the right track.

- mathmate

Gimme a minute, I have to flip the pages.
You can go get a glass of water, you talked a lot! lol

- pooja195

XD

- mathmate

Heard about compound inequalities? (6.5)

- pooja195

yes

- mathmate

Have you done interval notation?

- pooja195

no

- mathmate

Like (5,8(, or ]5, 8]

- pooja195

nope

- mathmate

They are usually used to solve compound inequalities.

- mathmate

Since you haven't done it, we'll describe by words.

- pooja195

You mean and and or ?

- pooja195

\[x+3<5~~or~~~5

- pooja195

like that?

- mathmate

Yes.

- mathmate

Can you try to solve
x<-5 or x>-4

- pooja195

They are already solved. -

- mathmate

Excellent!

- mathmate

Change to and:
Can you try to solve
x<-5 and x>-4

- pooja195

still the same :/ ?

- mathmate

No, I'll draw the number line.

- mathmate

|dw:1433387503479:dw|

- mathmate

Solution for or is shown in the graph.
But for and, there is no number that satisfies the and condition.
No number can be less than -5 and greater than -4 at the same time, so answer is "no solution".

- mathmate

How about solving:
x>4 and x<10
Use the number line if necessary

- mathmate

Feel free to show intermediate steps progressively.
You don't have to wait till you typed up everything.

- pooja195

|dw:1433387789387:dw|

- mathmate

The answer is correct graphically.
How would you write it mathematically?

- pooja195

x>4 and x<10

- mathmate

You can write it as
\(4\lt x\lt10\)
whe it is a range

- mathmate

But be careful about the difference between \(\le\) and \(\lt\) etc

- mathmate

Moving onto 6.6 unless you have questions.

- mathmate

Solve |x|=2

- pooja195

+2
-2

- mathmate

Very good. I see that your book does not write the answer in set notation, for example: x={-2,2}

- mathmate

Neither does your teacher?

- pooja195

Nope

- mathmate

ok, no problem.
Solve |x|<2

- pooja195

x<2 and x>âˆ’2

- mathmate

Again, these two "and" inequalities make a range between -2 and +2 so we write -2

- pooja195

ok

- mathmate

solve |x-3| =5

- pooja195

xâˆ’3=5 x=8
xâˆ’3=âˆ’5 x=-2

- mathmate

So x=8 or -2
Good, you know your stuff! This is encouraging! what do you think?

- pooja195

Its easy :)

- mathmate

Even better!

- mathmate

now solve |2x-7| -3=6

- pooja195

add 3 to both sides
2x-7=9 x=8
2x-7=-9 x=-1

- mathmate

Very good! It's still easy?

- pooja195

yes

- mathmate

We move onto abs. inequalities.

- mathmate

solve |4-3x| -2 \(\le\) 7

- pooja195

Do you want me to show the work?

- mathmate

Not in this case, will save some time.

- pooja195

\[\huge~x \ge \frac{ -5 }{ 3} ~~~and ~~~x \le \frac{ 13 }{ 3 }\]

- mathmate

Can you check if it is x>=13/3

- pooja195

âˆ’3x+4âˆ’4â‰¥âˆ’9âˆ’4 I subtracted 4 from both sides
âˆ’3xâ‰¥âˆ’13 i divided and flipped the signs
it should be \[\huge~\frac{ 13 }{ -3 }\]

- mathmate

I have 4-3x <= -9 (second case)
-3x <= -13
3x >= 13
so x>=13/3
do you agree?

- pooja195

oh right the negatives cancel

- mathmate

yes, they do (after flipping sign)

- pooja195

and then u flipped .-.

- mathmate

xD

- pooja195

LOL

- mathmate

Looks like you're good till 6.7 (perhaps as you expected).

- mathmate

Sorry, didn't achieve 2 chapters, not even one, but we're efficient.

- mathmate

No dull moment. :)

- pooja195

:)

- mathmate

If we have time later on, we may go back to word problems.

- pooja195

NO

- mathmate

Sure, we'll continue with 6.8 then.

- mathmate

The NO is a tell-tale sign! :)

- pooja195

-_-

- pooja195

\(\color{blue}{\text{Originally Posted by}}\) @mathmate
If we have time later on, we may go back to word problems.
\(\color{blue}{\text{End of Quote}}\)
NONONONONONON!!!

- mathmate

Not today. Want to do word problems when you're fresh. We'll move on to graphics in 2 variables. p.367

- mathmate

Can you tell me in a few words about graphing linear inequalities in two variables?

- pooja195

:/

- mathmate

Use words like region, feasible, vertical, horizontal lines, etc

- mathmate

ok, so I'll do it.

- mathmate

... another time! :)

- pooja195

yes :)

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