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,
Difference of two squares:

((root8)x-(root18)y)((root8)x+(root18)y)
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
(4x - 6y) (2x + 3 y)
Hi

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

And that would be the right answer.

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)]
Quote by jhardcore
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
Quote by jhardcore

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.
Quote by jhardcore

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)
Awesome, thanks a bunch guys
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

Quote by jhardcore
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

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.
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Sorry but that didn't really make sense
Quote by jhardcore
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!
Thanks man. Its so hard to understand though.
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:

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

OR

x = 7 - SQRT (49 - 8)
-------------------
2``````
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Quote by porcupineoris
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

Quote by porcupineoris
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,
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.
Quote by ColdbulleT
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
Quote by jhardcore
Edit: Go to bottom post please!

Quote by porcupineoris
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?
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dude, im a civil engineer.. i dun do math
...even when i'm sleepin' i'm keepin' it real.
divide by zero.