₦airaland Forum

benbuks: ok...you"ll have to go back to your basics of complex numbers

rationally , negative roots are invalid e.g ± √-1 , ±√-2 ,±√-3 e.t c.

solving x² + 1 =0 ..........(*)

produces complex roots , however mathematicians have worked severally on such irrational roots , which later discovered & agreed that solution to say (*) =±√-1

we now use ' i ' or ' j ' = ±√-1 ..........(**)

hence solution of (*) ,x = ± i

having believed that x=± i

=>i² =-1 .......( from (**) )

i³ = i(i²) =i(-1) =-i

i⁴ = (i² ) (i² ) = (-1)(-1)=1

i⁵ = (i⁴ ) i = (1)i = i

we could observe that powers of ' i ' in .

2 ,6 , 10 ,14 . . . =-1

3 ,7 , 11 ,15 . . . =-i

4 , 8 , 12 , 16 . . . = 1

5 , 9, 13 , 17 , . . . =i

& their multiples in that order , gives same result .
.

benbuks: --------SOLUTION--------------
(a)

= 3(i¹⁰⁴ )i + 5(i⁷² )i -8(i² )²⁵ -2(i¹¹⁸ ) i

=>3i(1) +5i(1) - 8(-1) -2i(-1)

= 10i +8

(b)

by difference of two squares.
we have
[(3i-5)² ]² - [(2-i)² ]²

=>(9i² -30i+25)² -(4-4i+i²)²

=>(16-30i)² - (3-4i)²

=> (256-960i +900i² )-(9-24i+16i² )
=>-544-960i +7 +24i

hence we have ,

-537 -36i

efficiencie: Let the matrix A with element a(ij) be defined as:
A=[row(1)(a(11) a(12) a(13) ...a(1m), row(2)(a(21) a(22) a(23) ...a(2m), ... row(n)(a(n1) a(n2) a(n3) ...a(nm)]

Hence if d determinant:
|row(1)(x y z), row(2)(a b c), row(3)(1 2 3)|=5
Then
|row(1)(1 2 9), row(2)(a b 3c), row(3)(x y 3z)|
Multiple of a column
=3. |row(1)(1 2 3), row(2)(a b c), row(3)(x y z)|
Interchange of rows
=-3. |row(1)(x y z), row(2)(a b c), row(3)(1 2 3)|
=-3(5)
=-15

hence
|row(1)(1 2 9), row(2)(a b 3c), row(3)(x y 3z)|=-15

I hope i'm ryt!!!

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Not my fault I was sick during d exams...Damn it

benbuks: my boss @ dej4u has justice already...BUT....

given ax² +bx + c =0

by definition ,

the sum of roots , A+B = -b/a

the product of roots , AB =c/a

but here we are given new roots build a quadratic equation with ,

A-2B & B-2A

now for sum , we have (A-2B) +(B-2A) =A-2B+B-2A
= (A+B) -2(A+B)

recall that A+B =- b/a
=> -b/a -2(-b/a)
= (-b+2b)/a = b/a

& product , (A-2B)(B-2A) =AB-2A² -2B² +4AB

=>

5AB -2(A² + B²)

but A² +B² =(A+B)² -2AB

=> 5AB -2[ ( A+B)² -2AB ]

=> 5c/a -2[ (b/a)² -2c/a]

=> 5c/a -2b² /a² + 4c/a

=> 9c/a - 2(b/a)²

using the general model for building quadratic equations , given roots ,

x² - (sum of roots ) x + product of roots =0

=> x² -(b/a)x +[ 9c/a -2(b/a)² ] =0

simplify further …

hope you get .?

dejt4u: the roots of the given eqn are, A nd B,
then, sum of the root, A+B=-b/a,
Product, AB= c/a,

for the required eqn,
roots are (A-2B) and (B-2A),
sum = (A-2B)+(B-2A) = -(A+B) = -(-b/a) = b/a,
product = (A-2B)(B-2A) = AB-2(A²+B²)+4AB = 5AB - 2(A²+B²)
=9AB-2(A+B)²
=[s]9(c/a) -2(-b/a) = (9c+2b)/a..

But the quadratic equation is of the form,
x² - (sum of the roots)x + (product of the roots)= 0
using the values aboves,

the required eqn is therefore,
x² - b(x)/a + (9c+2b)/a = 0,
ax² - bx + (9c+2b) = 0[/s]

modifications.,
product = (A-2B)(B-2A) = AB-2(A²+B²)+4AB = 5AB - 2(A²+B²)
=9AB-2(A+B)²
=9(c/a) -2(-b/a)² = (9ac-2b)/a²..

But the quadratic equation is of the form,
x² - (sum of the roots)x + (product of the roots)= 0
using the values aboves,

the required eqn is therefore,
x² - b(x)/a + (9ac-2b)/a² = 0,
multiplying all by a²
(ax)² - abx + (9ac-2b) = 0...QED
sorry for the previous mistake.. Tnx to oga benbuks

safarigirl: haba! Ghana has one bronze medal now, dem never try?

dejt4u: the roots of the given eqn are, A nd B,
then, sum of the root, A+B=-b/a,
Product, AB= c/a,

for the required eqn,
roots are (A-2B) and (B-2A),
sum = (A-2B)+(B-2A) = -(A+B) = -(-b/a) = b/a,
product = (A-2B)(B-2A) = AB-2(A²+B²)+4AB = 5AB - 2(A²+B²)
=9AB-2(A+B)²
=9(c/a) -2(-b/a) = (9c+2b)/a..

But the quadratic equation is of the form,
x² - (sum of the roots)x + (product of the roots)= 0
using the values aboves,

the required eqn is therefore,
x² - b(x)/a + (9c+2b)/a = 0,
ax² - bx + (9c+2b) = 0 ...QED

efficiencie: Given LOGaA•LOGbB=LOGbA•LOGaB

Starting frm d RHS:
LOGaA•LOGbB
=(LogA/Loga).(LogB/Logb)
=(LogA/Logb).(LogB/Loga)
=LOGbA•LOGaB

Which is the LHS

Nb LOGxX=LogX/Logx

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I was prepared ..... with my generator and fuel

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