Ortarico:

Yes you're right, cause with the options it's easier to work out what/how the examiner wants the equation to be.

doubleDx:

A little error there Chief Ortarico

=> log25^x₅ - log1/25₅
=> log(5)^{^2(x)}₅ - log(25)^(-1)₅
=> (2x)log5₅ - log(5)^2(-1)₅
=> (2x)log5₅ - (-2)log(5)₅
Since log5₅ = 1
=> (2x)log5₅ - (-2)log(5)₅
=> 2x .(1) + 2
=> 2x + 2
=> 2(x + 1)

So the final answer should => 2(x + 1)

Omosivie,:
Pls help me solve this: 1/root 2 -1 minus 1/root 2 +1. Thanks

Ortarico:

NB: _/ is for the root
1/(_/2 - 1) - 1/(_/2 + 1)
The LCM is (_/2 - 1)(_/2 + 1):
=> (_/2 + 1) - (_/2 - 1)/(_/2 - 1)(_/2 + 1)
Clearing the bracket we'll get:
=> _/2 + 1 - _/2 + 1/_/4 + _/2 - _/2 - 1

=> 1 + 1/2 - 1
=> 2. . . . . . . .ans

Omosivie,:
Pls help me out with this two simultaneous equations:
1, xy=-3; x+y=2
2, x+2y=0; x^2+y^2=20

Omosivie,:
Pls help me out with this two simultaneous equations:
1, xy=-3; x+y=2
2, x+2y=0; x^2+y^2=20

Ortarico:

Let's go. . . .
1. xy = -3. . . . . .eqn(i)
x + y= 2. . . . . . .eqn(ii)
from eqn(ii):
x= 2 - y. . . . . .eqn(iii)
substitute eqn(iii) in2 eqn(i):
xy= -3
(2 - y) y= -3
2y - y²= -3
Re-arrangng gives:
y² - 2y - 3= 0
(y - 3)(y + 1)
y= 3 or -1
:. x= -1 or 3
To check:
xy= -3; (-1*3)= -3
x + y= 2; (-1 + 3)= 2

2. x + 2y= 0. . . . . .eqn(i)
x² + y²= 20. . . . .eqn(ii)
from eqn(i):
x + 2y= 0
x= -2y. . . . . . .eqn(iii)
substitute eqn(iii) in2 eqn(ii):
x² + y²= 20
(-2y)² + y²= 20
4y² + y²= 20
5y²= 20
y²= 4
y= 2
substitute y= 2 in2 eqn(i):
x + 2y= 0
x + 2(2)= 0
x + 4= 0
x= -4
To check:
x + 2y= 0; -4 + 2(2)= 0
x² + y²= 20; (-4)² + (2)²= 20

Leebliss13: wat r d opti0ns?

Omosivie,:
Thanks, but the answer is not the same with my textbook's answer. According to my textbook, the answer for number one is -3 or 1 and the answer for number two is (4,2) or (-4,2). I don't understand dis textbook and i'm totally fed up.

Ortarico:

But that's it now.
At no 1 x is (-3 , 1) then y is (-1 , 3) and at no 2 x is -4, then y is 2. To check that you saw was just to confirm. That's what I got sister

Omosivie,:
Thanks, I definately need glasses

Calculusf(x):
Remember this bro,the square root of a natural number is + or -...so,from where you got y^2=4...therefore y=+ or - 2...therefore y=+2 or -2...from the first equation x+2y=0...substitute y=-2...therefore x+2(-2)=0...x-4=0...x=4...substitute y=2...x+2y=0...x+2(2)=0...x+4=0...x=-4...therefore,when y=-2,x=4 and when y=2,x=-4

Omosivie,:
I hope i'm not bothering u guys with these "easy" questions but can u please help me out with them
1, m^3 + mn^2-nm^2-n^3/m^4-n^4
2, -12a^2x/-8a^2b^2

Ortarico: 2. Please re-type, there's a mistake or I don't get you well.

Ortarico:

1. m³ + mn² - nm² - n³/m⁴ - n⁴
m³ + n³ + mn² - nm²/m⁴ - n⁴
(m - n)³ + mn(n - m)/(m - n)⁴
mn(n - m)/(m - n) . . . . .ans

Omosivie,:
I hope i'm not bothering u guys with these "easy" questions but can u please help me out with them
1, m^3 + mn^2-nm^2-n^3/m^4-n^4

doubleDx:

{m³ + mn² - nm² - n³}/(m⁴ - n⁴)

Simplifying m⁴ - n⁴ yields => (m²^2 - n²^2 => (m² - n²)(m² + n²)...difference of 2 squares.

Rearranging the numerator & factorising yields =>
=> mn² - nm² + m³ - n³
=> mn(n - m) - (n³ - m³)
Now, (n³ - m³) can be rewritten as => (n - m)³ + 3mn(n - m) - difference of two cubes, which makes the numerator => mn(n - m) - {(n - m)³ + 3mn(n - m)}

Putting the numerator & denominator together yields:

=>> [mn[b](n - m)[/b] - {(n - m})³ + 3mn((n - m))]/(m² - n²)(m² + n²)

Factorising (n - m) from the above yields=>

=>> [(n - m)][mn - {(n - m)² + 3mn}]/(m² - n²)(m² + n²)

=>> [n - m][mn - {(n² - 2mn + m² + 3mn)]/(m² - n²)(m² + n²)

=>> [n - m][mn - {(n² + m² + mn)]/(m² - n²)(m² + n²)

=>> (n - m)(-n² - m²)/(m² - n²)(m² + n²)

=>> (n - m)(-1)(n² + m²)/(m² - n²)(m² + n²)
=>> (m - n)(n² + m²)/(m² - n²)(m² + n²)


=>> (m - n)/(m² - n²)

=>> (m - n)/(m - n)(m + n)

=>> 1/(m + n)

Hence,

{m³ + mn² - nm² - n³}/(m^4 - n^4) = 1/(m + n)

Omosivie,:
Wow!!!! This is so complex(one of the reasons why I hate maths). Could this be further maths?? I just pray I'm not given this question in my upcoming exam cos it's too long, difficult and complex. Thanks for solving it.

Omosivie,:
-12a^2 x / -8a^2 b^2

doubleDx:

{m³ + mn² - nm² - n³}/(m⁴ - n⁴)

Simplifying m⁴ - n⁴ yields => (m²^2 - n²^2 => (m² - n²)(m² + n²)...difference of 2 squares.

Rearranging the numerator & factorising yields =>
=> mn² - nm² + m³ - n³
=> mn(n - m) - (n³ - m³)
Now, (n³ - m³) can be rewritten as => (n - m)³ + 3mn(n - m) - difference of two cubes, which makes the numerator => mn(n - m) - {(n - m)³ + 3mn(n - m)}

Putting the numerator & denominator together yields:

=>> [ mn((n - m) - {(n - m})³ + 3mn((n - m))]/(m² - n²)(m² + n²)

***** Factorising (n - m) from the numerator above yields=>

=>> [(n - m)][mn - {(n - m)² + 3mn}]/(m² - n²)(m² + n²)

=>> [n - m][mn - {(n² - 2mn + m² + 3mn)]/(m² - n²)(m² + n²)

=>> [n - m][mn - {(n² + m² + mn)]/(m² - n²)(m² + n²)

=>> (n - m)(-n² - m²)/(m² - n²)(m² + n²)

=>> (n - m)(-1)(n² + m²)/(m² - n²)(m² + n²)
=>> (m - n)(n² + m²)/(m² - n²)(m² + n²) *****


=>> (m - n)/(m² - n²)

=>> (m - n)/(m - n)(m + n)

=>> 1/(m + n)

Hence,

{m³ + mn² - nm² - n³}/(m⁴ - n⁴) = 1/(m + n)

Ortarico:

Respect to you @ general doubleDx. I supposed to have used the law of cubic and quadratic expression but for my hurry I missed it.
Please help explain more at the five starred phase of your work, thanks general!

doubleDx:

[ mn (n - m) - {(n - m)³ + 3mn (n - m)}]/(m² - n²)(m² + n²)

If you look at the numerator of the above expression carefully, you will realize that (n - m) is a common factor/coefficient @ mn (n - m) , (n - m)³ and 3mn (n - m)

So that if (n - m) is replaced with say Q, the numerator would be => [ mn Q - {(Q)³ + 3mn(Q)}] which reduces the numerator to =>

mnQ - {Q³ + 3mnQ}
mnQ - Q³ - 3mnQ

Now, if Q is factorized from the above, it becomes

=> Q(mn - Q² - 3mn)
=> Q(-2mn - Q²)
=> -Q(2mn + Q²)

Remember that Q = (n - m), so that -Q = (m - n), putting back the values of Q & -Q into the numerator's expression, yields =>

(m - n)(2mn + (n - m)²)
(m - n)(2mn + n²+ m² - 2mn)
(m - n)(n² + m²)

Putting the simplified numerator back into the expression yields =>

=>> (m - n)(n² + m²)/(m² - n²)(m² + n²)

=>> (m - n)/(m² - n²)

=>> (m - n)/(m - n)(m + n)

=>> 1/(m + n)

I hope this is clearer...

doubleDx:

It's not Further Maths lol, the solution is not as long as it appears, though my explanation is, so you could understand better. It's just a matter of understanding two expressions and using them at the right time =>

1. That A² - B² = (A - B)(A + B) - difference of two squares
2. That A³ - B³ = (A - B)³ - 3AB(A - B) - difference of two cubes

So that A⁴ - B⁴ = (P² - Q²) => (P - Q)(P + Q), such that =>> P = A² and Q = B².

Understanding these expressions and knowing when to apply/fix them in when you factorize makes the question weak and easy to solve.

I hope you catch my drift... so, you see? It's easy, don't be discouraged god sister!