brackets within brackets

• Oct 4th 2009, 05:51 AM
wolfhound
brackets within brackets
hi how would I work out the following?
(3(2^2 -2)^2 -10)^2

which brackets do I calculate first please?
• Oct 4th 2009, 06:15 AM
e^(i*pi)
Quote:

Originally Posted by wolfhound
hi how would I work out the following?
(3(2^2 -2)^2 -10)^2

which brackets do I calculate first please?

1. $\displaystyle 2^2-2 = {\color{red}2}$
2. $\displaystyle {\color{red}2}^2 = {\color{blue}4}$
3. $\displaystyle 3 \times {\color{blue}4} = {\color{green}12}$
4. $\displaystyle {\color{green}12} - 10 = 2$
5. $\displaystyle 2^2 = 4$
• Oct 4th 2009, 06:48 AM
wolfhound
Thanks (Clapping)
• Oct 4th 2009, 07:03 AM
e^(i*pi)

• Oct 4th 2009, 07:24 AM
wolfhound
value of x
find value of x if x-2 =5x + 14

I havent learned how to do these yet I just want to find out how and then I will catch on
Thanks
• Oct 4th 2009, 07:25 AM
Quote:

Originally Posted by wolfhound
find value of x if x-2 =5x + 14

I havent learned how to do these yet I just want to find out how and then I will catch on
Thanks

HI

$\displaystyle x-2=5x+14$

$\displaystyle x-5x=14+2$

$\displaystyle -4x=16$

$\displaystyle x=-4$
• Oct 4th 2009, 07:45 AM
wolfhound
Quote:

HI

$\displaystyle x-2=5x+14$

$\displaystyle x-5x=14+2$

$\displaystyle -4x=16$

$\displaystyle x=-4$

cool thanks (Rofl)
• Oct 4th 2009, 07:51 AM
wolfhound
another find x
find value of x if
2x-1/x+2 = 3

and
is 5 a solution of x^2 + 3x-9=(x-4)(x+21)??
If so please explain how I do it
Thanks!!!!!
• Oct 4th 2009, 07:53 AM
wolfhound
actually is x =7 the first one...
• Oct 4th 2009, 08:10 AM
Quote:

Originally Posted by wolfhound
find value of x if
2x-1/x+2 = 3

and
is 5 a solution of x^2 + 3x-9=(x-4)(x+21)??
If so please explain how I do it
Thanks!!!!!

HI again .

If you need any more help , its better to start a new thread . Well , i am going to answer this . If you have any more q's , start a new onw

$\displaystyle \frac{2x-1}{x+2}=3$

$\displaystyle 2x-1=3(x+2)$

$\displaystyle 2x-1=3x+6$

$\displaystyle x=-7$

$\displaystyle x^2+3x-9=(x-4)(x+21)$

$\displaystyle x^2+3x-9=x^2+17x-84$

$\displaystyle 3x-9=17x-84$

$\displaystyle 14x=75$

$\displaystyle x=75/14$

So 5 is not a solution .
• Oct 4th 2009, 08:14 AM
wolfhound
Quote:

HI again .

If you need any more help , its better to start a new thread . Well , i am going to answer this . If you have any more q's , start a new onw

$\displaystyle \frac{2x-1}{x+2}=3$

$\displaystyle 2x-1=3(x+2)$

$\displaystyle 2x-1=3x+6$

$\displaystyle x=-7$

$\displaystyle x^2+3x-9=(x-4)(x+21)$

$\displaystyle x^2+3x-9=x^2+17x-84$

$\displaystyle 3x-9=17x-84$

$\displaystyle 14x=75$

$\displaystyle x=75/14$

So 5 is not a solution .

Thanks so much for your help I will start a new thread when I have more questions... (Bow)(Happy)
• Oct 4th 2009, 08:18 AM
e^(i*pi)
You seem to have the wrong sign for both of them

$\displaystyle \frac{2(7)-1}{7+2} = \frac{13}{9} \neq 3$

by inspection $\displaystyle x = -7$

-----------------

For the 2nd one $\displaystyle 31 \neq 26$ so 5 is not a solution

$\displaystyle x^2+3x-9 = x^2+17x-84$

$\displaystyle -14x = - 75$