Ask Singapore Homework?
Upload a photo of a Singapore homework and someone will email you the solution for free.
Question
secondary 3 | A Maths
One Answer Below
Anyone can contribute an answer, even non-tutors.
Pls help me with this qn
Pls reply asap thx
Date Posted: 3 years ago
Views: 237
consider p^2 ...
regardless of whether p is positive , 0 or negative,
result of p^2 is greater than or equal to 0.
similarly for any p^n where n is even.
for the given equation, since both terms are even power, the only possible solution is ...
(2x-1) = 0 and (y+2) = 0
x=(1/2) and y = -2
x-y = 5/2
regardless of whether p is positive , 0 or negative,
result of p^2 is greater than or equal to 0.
similarly for any p^n where n is even.
for the given equation, since both terms are even power, the only possible solution is ...
(2x-1) = 0 and (y+2) = 0
x=(1/2) and y = -2
x-y = 5/2
Given the power of 4 which is even, (2x-1)^4 and (y+2)^4 must both be greater than or equal to 0. Since the addition of 2 positive numbers cannot equal to 0, both (2x-1)^4 and (y+2)^4 are equal to 0. Hence x = 1/2 and y = -2, and x-y = 5/2
See 1 Answer
done
{{ upvoteCount }} Upvotes
clear
{{ downvoteCount * -1 }} Downvotes
If X and y are real
Then firth piers are positive
The only way the sum are zero are both are zero
Then you could get x and y easily
Then firth piers are positive
The only way the sum are zero are both are zero
Then you could get x and y easily
Date Posted:
3 years ago
Thx
360 Tutoring Program