Expert: Ahmed Salami - 6/11/2004

Question
please can you help me with equations, thank you so much


#1

standard form a+bi

2+3i/1-i

#2
2x^2-4x+1=0

#3
find the real solution
square root 3-x+x^2=x-1

#4
square root 3x+7+square root x+2=1

Answer
Hi jacira,
Sorry for the delay.
Lets go through it as easy as possible.
(1) The first thing you do here is rationalize as we do when we tackle surds. We therefore introduce the complex conjugate of 1-i which is 1+i
therefore,
2+3i/1-i = (2+3i/1-i).(1+i/1+i)
        = (2+3i)(1+i)/(1-i)(1+i)
        = (2+2i+3i-3)/(1+1)
        = (-1+5i)/2
        = (-1/2) + (5/2)i

(2)For 2x^2-4x+1 = 0
we have,
x = |4 + or - sqrt(4^2 - 4.2.1)| /2.2
 = |4 + or - sqrt(16 - 8)| /4
 = |4 + or - sqrt8| /4
 = |4 + or - sqrt4.sqrt2| /4
 = |4 + or - 2sqrt2| /4
 = 1 + or - sqrt2/2

(3)sqrt(3-x+x^2) = x-1
squaring both sides,
3-x+x^2 = (x-1)^2
3-x+x^2 = x^2 - 2x + 1
rearranging gives
2x - x = 1 - 3
x = -2

(4)sqrt(3x+7)+ sqrt(x+2) = 1
squaring both sides,
(3x+7)+ 2sqrt(3x+7)(x+2)+ (x+2) = 1
(4x+9)+ 2sqrt(3x^2 + 13x + 14) = 1
2sqrt(3x^2 + 13x + 14) = 1 - 4x - 9
2sqrt(3x^2 + 13x + 14) = -4x - 8
sqrt(3x^2 + 13x + 14) = -2x - 4
squaring again,
3x^2 + 13x + 14 = 4x^2 + 16x + 16
rearranging gives
x^2 + 3x + 2 = 0
by factorization,
(x+1)(x+2) = 0
Therefore x = -1 or -2

I hope this is helpful, and remember that the method of solution is always the most important.
Regards.