[madmaths]

Expand the brackets in the following expression
(5y-3)^2

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Factorise the following expressions

x^2 - x - 12 (2)


9y^2-25

I've several questions like this if someone could let me know what to do with this one then maybe I could get an idea of what to do with the rest. Thanks!

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solve the equation.

9/(x=7)= 4/(x-3)

[Unknown008]

1. For this one, remember that:



2. (a) Let's say the factorised expression is (x - a)(x - b), when you expand it out, you get:



Now, (a + b) is 1 from the given and ab = -12

can you guess what a and b can be?

(b) Use the difference of two squares, that is, if you have:



3. I guess you mean:



?

Cross multiply, then move all the terms in x and the terms without x on separate sides of the equation. Last, divide by the coefficient of x.

Post what you get! :)

[madmaths]

thanks going by that the answer I work out is x=-1/5 .

[Unknown008]

For the last question?

Hm... I'm getting something else.

Cross-multiplying:



Expanding:



Move around:







Is that okay now? :)

[madmaths]

Hi yes sorry it was the last question.

I cross multiplied:
9*(x-3)=4*(x-7) I notice you kept the + sign in thesecond bracket, I thought it became negative.
so then I worked out
9x-27=4x-28
Then I collected the variable x on the left side and the rest on the right
9x-4x=27-28
5x=-1
then I divided the right side with the coefficient of x
x=-1/5

[madmaths]

For question 1.
I get
(5y-3)^2=(5y-3)(5y-3)
(5y-3)(5y-3)=(5y)^2-(2.3.5y)+3^2
(5y)^2-(2.3.5y)+3^2=25y^2-30y+9

[Unknown008]

Oh, instead of '=', you meant '-', okay, I thought it was '+' since it's the sign usually replaced by '='.

Okay, last one is good :)

First one is good too :)

[madmaths]

Thanks for all your help, I've worked out the other two, hopefully they are right!

I've worked out 9y^2-25 to be (3x-5((3x=5)

and x^2-x-12 to be (x-)(x=3)

[Unknown008]

Oh well, that's not too easy to understand.

9y^2 - 25 = (3y + 5)(3y - 5)

x^2 - x - 12 = (x - 4)(x + 3)

:)

[RPVega]

(5y-3)^2 = (5y-3)(5y-3)=25y - 15y -15y + 9 = 25y - 30y + 9
Remember the word "FOIL": First, Outer, Inner, Last
Multiply the First terms of the two binomials: 5y * 5y = 25y
Multiply the Outer terms of the two binomials: 5y * -3 = -15y
Multiply the Inner terms of the two binomials: -3 * 5y = -15y
Multiply the Last terms of the two binomials: -3 * -3 = 9
Finally, add all the terms together: 25y + (-15y) + (-15y) + 9 = 25y -30y + 9

9/(x-7) = 4/(x-3)
4(x-7) = 9(x-3)
4x - 28 = 9x - 27
-28 + 27 = 9x - 4x
-1 = 5x
x = -1/5

[RPVega]

Sorry, I made a boo,boo. (5y-3)^2 = 25y^2-30y+9
(5y-3)^2 = (5y-3)(5y-3)
Remember the word "FOIL": First, Outer, Inner, Last
Multiply the First terms in the two binomials: 5y*5y=25y^2
Multiply the Outer terms in the two binomials: 5y*(-3)=-15y
Multiply the Inner terms in the two binomials: (-3)*5y=-15y
Multiply the Last terms in the two binomials: (-3)*(-3)=9
Next, add the results of all the above together: 25y^2+(-15y)+(-15y)+9=
25y^2-30y+9

Here's a shortcut to the above:

When taking the square of a binomial, (a+b)^2, take the square of the first
term, a^2, plus the product of the two terms multiplied by 2, 2ab, plus
the square of the last term, b^2. The result is: (a+b)^2=a^2 + 2ab + b^2.
Try this with the above example: (5y-3)^2. With this shortcut, you can arrive
at the answer in ONE step: (a+b)^2=(5y+(-3))^2=(5y)^2-(5y*(-3)*2)+(-3)^2=
25y^2-30y+9, where a=5y; and b=(-3). Good luck! The U.S. needs more engineers and math teachers!

[RPVega]

9y^2-25=(3y+5)(3y-5)
Here's a shortcut:
(ay+b)(ay-b)=(ay)^2-b^2. This applies when multiplying two binomials where the b terms
are of opposite signs in the two binomials. In the above example, a=3; and b=5.
Good luck!