#1

Edit: Go to my post about 8 please!

Hey guys, I have to do some maths. I am pretty sure I know what to do, but I just need some confirmation and a bit of help.

all 'x' mean the algebraic symbol x, not multiply.

Small 2's mean squared

the questions are;

4(5x-3)2

I assume I would go;

4 (5x-3)(5x-3)

4 [25x2 -15x -15x + 9]

4 [25x2 -30x + 9]

100x2 - 120x + 36

-(x-2y)(3x+5y)

I assume you would go;

-3x2-5xy+6xy-10y2

-3x+xy-10y2

8x2-18y2

I don't how to do this

Help would be appreciated,

Thanks

Hey guys, I have to do some maths. I am pretty sure I know what to do, but I just need some confirmation and a bit of help.

all 'x' mean the algebraic symbol x, not multiply.

Small 2's mean squared

the questions are;

__Expand the brackets:__4(5x-3)2

I assume I would go;

4 (5x-3)(5x-3)

4 [25x2 -15x -15x + 9]

4 [25x2 -30x + 9]

100x2 - 120x + 36

__Expand the brackets:__-(x-2y)(3x+5y)

I assume you would go;

-3x2-5xy+6xy-10y2

-3x+xy-10y2

__Factorise:__8x2-18y2

I don't how to do this

Help would be appreciated,

Thanks

*Last edited by jhardcore at Apr 3, 2008,*

#2

Difference of two squares:

((root8)x-(root18)y)((root8)x+(root18)y)

((root8)x-(root18)y)((root8)x+(root18)y)

#3

This is the MyImmortal law of math: If it has the word math, or is in any way related to math, it shall be dropped, and if solution 1 can not happen it shall be nap time.

/my 2 cents

/my 2 cents

#4

(4x - 6y) (2x + 3 y)

#5

(4x - 6y) (2x + 3 y)

And that would be the right answer.

#6

your first one is correct

the second one should have been done in the same way as the first

as in, expand your brackets, and then mulitply through by -1

this is how i did the third one

=8x^2 - 18y^2

=2(4x^2 - 9y^2)

=2[(2x-3y)(2x+3y)]

the second one should have been done in the same way as the first

as in, expand your brackets, and then mulitply through by -1

this is how i did the third one

=8x^2 - 18y^2

=2(4x^2 - 9y^2)

=2[(2x-3y)(2x+3y)]

#7

Expand the brackets:

4(5x-3)2

I assume I would go;

4 (5x-3)(5x-3)

4 [25x2 -15x -15x + 9]

4 [25x2 -30x + 9]

100x2 - 120x + 36

Correct

Expand the brackets:

-(x-2y)(3x+5y)

I assume you would go;

-3x2-5xy+6xy-10y2

-3x+xy-10y2

Nearly, its: -3x^2+xy+10y^2 You just got the wrong sign for the y^2 coefficient.

Factorise:

8x2-18y2

Take out the common factor of 2: 2(4x^2-9y^2)

Now in the brackets you have a difference of two squares, which gives:

2(2x+3y)(2x-3y), which as someone else has said is the same as:

(4x+6y)(2x-3y)=(2x+3y)(4x-6y)

#8

Awesome, thanks a bunch guys

#9

Sorry to bring this back, but I need one more bit of help!

Factorise:

x^2 - 7x + 2

I know I am meant to start with (x )(x ) and find 2 numbers that add to make -7 and multiply to make 2, however I don't think such numbers exist

Thanks in advance.

Factorise:

x^2 - 7x + 2

I know I am meant to start with (x )(x ) and find 2 numbers that add to make -7 and multiply to make 2, however I don't think such numbers exist

Thanks in advance.

#10

Sorry to bring this back, but I need one more bit of help!

Factorise:

x^2 - 7x + 2

I know I am meant to start with (x )(x ) and find 2 numbers that add to make -7 and multiply to make 2, however I don't think such numbers exist

Thanks in advance.

In that case, in order to solve to find x, you would use the quadratic equation:

```
x = -b ± SQRT (b² - 4ac)
----------------------
2a
```

based on the form ax² + bx + c.

#11

Sorry but that didn't really make sense

#12

Sorry but that didn't really make sense

Ok, you have the general quadratic equation ax^2+bx+c=0, where a, b, c are real numbers.

Now, begin by dividing by a: x^2+(b/a)x+c/a=0

Then complete the square: (x+b/2a)^2+c/a-(b/2a)^2=x^2+(b/a)x+c/a=0

Take constants over (and simplify to a single fraction): (x+b/2a)^2=b^2-4ac

then equate to x: x=[-b+rt(b^2-4ac)]/2a

(remember the positive and negative square roots!)

This gives a formula for the roots of any quadratic equation, just plug in the numbers!

#13

Thanks man. Its so hard to understand though.

#14

Okay, if you arent able to factorise normally, you have to use the quadratic equation. The general form of a quadratic formula is ax² + bx + c, so for your equation;

a=1

b=-7

c=2

You then put this into the quadratic formula x = -b ± SQRT...etc etc

So you have the eqation:

a=1

b=-7

c=2

You then put this into the quadratic formula x = -b ± SQRT...etc etc

So you have the eqation:

```
x = 7 + SQRT (49 - 8)
-------------------
2
OR
x = 7 - SQRT (49 - 8)
-------------------
2
```

#15

Ok, you have the general quadratic equation ax^2+bx+c=0, where a, b, c are real numbers.

Now, begin by dividing by a: x^2+(b/a)x+c/a=0

Then complete the square: (x+b/2a)^2+c/a-(b/2a)^2=x^2+(b/a)x+c/a=0

Take constants over (and simplify to a single fraction): (x+b/2a)^2=b^2-4ac

then equate to x: x=[-b+rt(b^2-4ac)]/2a

(remember the positive and negative square roots!)

This gives a formula for the roots of any quadratic equation, just plug in the numbers!

Oh and another one

```
x = 7 + SQRT (49 - 8)
-------------------
2
OR
x = 7 - SQRT (49 - 8)
-------------------
2
```

now I understand

ake constants over (and simplify to a single fraction): (x+b/2a)^2=b^2-4ac

then equate to x: x=[-b+rt(b^2-4ac)]/2a

(remember the positive and negative square roots!)

This gives a formula for the roots of any quadratic equation, just plug in the numbers!

Listen to him

*Last edited by Liger02 at Apr 3, 2008,*

#16

Thanks man. Its so hard to understand though.

Yeah, sorry man. Look at the overview section of http://en.wikipedia.org/wiki/Completing_the_square#Proof and it shows you what i wrote more clearly

#17

Completing the square would be more simpler if you don't know the quadratic formula. But once you know the quadratic formula, you just sub in the numbers and simplify.

#18

Completing the square would be more simpler if you don't know the quadratic formula. But once you know the quadratic formula, you just sub in the numbers and simplify.

They're the same thing- solving using the quadratic formula is the same (albeit faster) method as completing the square

#19

Edit: Go to bottom post please!

They're the same thing- solving using the quadratic formula is the same (albeit faster) method as completing the square

this is the post at the bottom and now?

#20

dude, im a civil engineer.. i dun do math

#21

divide by zero.