Question

Please solve it immediately for getting 30 points.....Please solve it now.

Answer 1

plz wait sorry to write in bad handwriting but I was in hurry please contact if you need any help

Answer 2

Heya.......


 4x² - 2x + (k-4) = 0 -----(1)



Roots are α and 1α 


α * 1/α = c/a 
1 = (k - 4) / 4 
4 = k - 4 
k = 8 


==================== 


Problem doesn't ask to find α, but I will anyway to show k = 8 is indeed valid. 


4x² - 2x + (k-4) = 0 
4x² - 2x + (8-4) = 0 
4x² - 2x + 4 = 0 
2 (2x² - x + 2) = 0 
2x² - x + 2 = 0 


x = [1 ± √(1-16)] / 4 
x = (1 ± i√15) / 4 


The zeros of polynomial are (1 + i√15) / 4 and (1 − i√15) / 4 
It may not look like it but these are actually reciprocals of each other: 


Let α = (1 + i√15) / 4 


1/α = 4 / (1 + i√15) 
..... = 4 (1-i√15) / ((1+i√15)(1-i√15)) 
..... = 4 (1-i√15) / (1-15i²) 
..... = 4 (1-i√15) / 16 
..... = (1 − i√15) / 4 


But this is just the other zero. 


So the two roots (1 + i√15) / 4 and (1 − i√15) / 4 are reciprocals of each other ..........


So when k = 8, we get zeros that are reciprocals of each other. 
In fact when a = c we always get zeros that are reciprocals of each other.

HOPE THIS WILL HELP U---:)

tysm..@Gozmit