NewStats: 3,265,115 , 8,185,715 topics. Date: Friday, 13 June 2025 at 02:19 PM g705b

Ortarico:

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

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 be => 2(x + 1) which is option B as solved by Omosivie

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)

Ok thanks general doubleDx

To find the sum of positive integers...1+2+3+...+n=n(n+1)/2...

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

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

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

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

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

wat r d opti0ns?

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

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

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

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.

Leebliss13: wat r d opti0ns?

No options

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.

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

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

Thanks, I definately need glasses

Omosivie,:
Thanks, I definately need glasses

Thanks too, it happens to everyone.

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

Calculusf(x):
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

All right, but if you look at it well, you'll see that I typed +/- 2, I know the rule boss.

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

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

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

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

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

-12a^2 x / -8a^2 b^2

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

The answer @ the back of my textbook is 1/m+n

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

{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)

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)

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 exam cos it's too long, difficult and complex. Thanks for solving it.

@ Ortarico, Calucus(fx) and DoubleDx. Pls help me out with the number 2 question

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 exam cos it's too long, difficult and complex. Thanks for solving it.

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!

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

Solution

-12a² x /-8a² b²

=> This is easy seeing as -4a² is a factor of the numerator and denomitor, cancelling out & reducing the expression to =>

For the numerator => -12a²/-4a² = 3x
For the denomitor => -8a² b²/-4a² = 2b²

Putting the simplified numerator and denomitor together yields =>

=> 3x/2b²

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)

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!

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!

[ 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)

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:

[ 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)

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...

Oh yes, thanks boss, one love!

You have really done a great job@general doubledx...

^

Thanks bruv!

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!

Thanks, I'll cram these algebraic identities/expressions.