Prove that for any positive numbers p and q: p + q = sqrt(4pq)

Watch
EmRep13
Badges: 13
Rep:
?
#1
Report Thread starter 1 month ago
#1
I did (p + q)^2 = p^2 + 2pq + q^2 g ≥ 4pq
so p^2 -2qr + q^2 ≥ 0 which is true since it's equal to (p-q)^2 and any number squared is ≥ 0.
Since (p - q)^2 + 4pq = (p + q)^2 and (p - q)^2 ≥ 0
(p + q)^2 ≥ 4pq so
p + q ≥ sqrt (4 pq)

The solution bank added the line p and q are both positive
so p > 0 and q > 0
Therefore, p + q > 0

Why is this necessary? Doesn't the line "...(p-q)^2 and any number squared is ≥ 0 " eliminate the need for it, since it shows that 4qr is the value that (p +q)^2 is equal or greater than?
0
reply
EmRep13
Badges: 13
Rep:
?
#2
Report Thread starter 1 month ago
#2
*bump* whoops
0
reply
zetamcfc
Badges: 19
Rep:
?
#3
Report 1 month ago
#3
(Original post by EmRep13)
I did (p + q)^2 = p^2 + 2pq + q^2 g ≥ 4pq
so p^2 -2qr + q^2 ≥ 0 which is true since it's equal to (p-q)^2 and any number squared is ≥ 0.
Since (p - q)^2 + 4pq = (p + q)^2 and (p - q)^2 ≥ 0
(p + q)^2 ≥ 4pq so
p + q ≥ sqrt (4 pq)

The solution bank added the line p and q are both positive
so p > 0 and q > 0
Therefore, p + q > 0

Why is this necessary? Doesn't the line "...(p-q)^2 and any number squared is ≥ 0 " eliminate the need for it, since it shows that 4qr is the value that (p +q)^2 is equal or greater than?
Nowhere in your proof do you use that p & q are positive....
0
reply
EmRep13
Badges: 13
Rep:
?
#4
Report Thread starter 1 month ago
#4
(Original post by zetamcfc)
Nowhere in your proof do you use that p & q are positive....
I don't see why you'd need to? Other than in the last bit just to state that it only applies if p + q are positive? But is that grounds to do so? Do you need to use it if the question states?
0
reply
thekidwhogames
Badges: 17
Rep:
?
#5
Report 1 month ago
#5
(Original post by EmRep13)
*bump* whoops
When they square rooted both sides of the inequality (p+q)^2 >= 4pq they used the fact that p,q are both positive (a,b>0 a^2>b^2 <==> a>b)) so you need to state this EmRep13
0
reply
_gcx
Badges: 21
Rep:
?
#6
Report 1 month ago
#6
(Original post by EmRep13)
I don't see why you'd need to? Other than in the last bit just to state that it only applies if p + q are positive? But is that grounds to do so? Do you need to use it if the question states?
(-2)^2 &gt; (-1)^2 but -2 &lt; -1, the problem comes when you go from (p + q)^2 \ge 4pq to p + q \ge \sqrt {4 pq}. (some constraint on p, q is required for this inequality to even make sense)
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Has your university communicated any last minute changes?

Yes, and they were what I was expecting (9)
17.31%
Yes, and they were not what I was hoping for (14)
26.92%
Yes, and they don't change my view on attending uni (3)
5.77%
No, I haven't received any communication (26)
50%

Watched Threads

View All
Latest
My Feed