# [SOLVED] Hard Algebra Questions

• May 17th 2006, 04:20 PM
ksle82
[SOLVED] Hard Algebra Questions
Simplify the followings:

A) (a^4 + b^4)/(a^2 +b^2)

B) (2 + 2i)/(3-2i), where i is the imaginary
• May 17th 2006, 04:58 PM
topsquark
Quote:

Originally Posted by ksle82
Simplify the followings:

A) (a^4 + b^4)/(a^2 +b^2)

B) (2 + 2i)/(3-2i), where i is the imaginary

For the way most would want the answer, A) is already as simple as it is going to get.

In B) we want to remove the radical in the denominator. So multiply top and bottom by the complex conjugate of the denominator:
$\frac{2+2i}{3-2i} \cdot \frac{3+2i}{3+2i}$

Since $i^2=-1$:
$=\frac{6+4i+6i+4i^2}{9+6i-6i-4i^2}=\frac{6+10i-4}{9+4}$

$=\frac{2+10i}{13}$

-Dan
• May 17th 2006, 05:27 PM
ticbol
Quote:

Originally Posted by ksle82
Simplify the followings:

A) (a^4 + b^4)/(a^2 +b^2)

B) (2 + 2i)/(3-2i), where i is the imaginary

Here are some ways:

A) (a^4 +b^4) / (a^2 +b^2) --------(1)

A.1) By long division, = (a^2 -b^2) +(2b^4)/(a^2 +b^2) -----------------------answer.

A.2) Get a factor (a^2 +b^2) from the numerator (a^4 +b^4):
Squaring the denominator,
(a^2 +b^2)^2 = a^4 +2(a^2)(b^2) +b^4
(a^2 +b^2)^2 = (a^4 +b^4) +2(a^2)(b^2)
(a^4 +b^4) = (a^2 +b^2)^2 -2(a^2)(b^2) ------(i)

Substitute that into (1),
(a^4 +b^4) / (a^2 +b^2) --------(1)
= [(a^2 +b^2)^2 -2(a^2)(b^2)] / (a^2 +b^2)
= (a^2 +b^2) -2(a^2)(b^2)/(a^2 +b^2) --------------answer.

=============================
B) (2 +2i) / (3 -2i) -----------------------(2)

Make the denominator a real number, muliply both numerator and denominator by the conjugate of the denominator, that is by (3 +2i):

= [(2 +2i)(3 +2i)] / [(3 -2i)(3 +2i)]
= [6 +4i +6i +4i^2] / [9 -4i^2]
Since i^2 = -1, then,
= [2 +10i] / [13]
= (1/13)(2 +10i)