Integers p,q,r satisfy p+q-r=-1 and p^2+q^2-r^2=-1 What is the sum of all positive values of p+q+r.
Answers
The correct data is:
p+q-r=1 and p^2+q^2-r^2=-1
Given:
Integers p,q,r satisfy p+q-r=-1 and p^2+q^2-r^2=-1
To find:
What is the sum of all positive values of p+q+r.
Solution:
From given, we have,
p+q-r=1 and p^2+q^2-r^2=-1
p+q-r=1 ⇒ p + q = 1 + r ......(1)
p^2+q^2-r^2=-1 ⇒ p^2 + q^2 = r^2 - 1 = (r + 1) (r - 1)
p^2 + q^2 = (p + q) (p + q - 2)
⇒ p^2 + q^2 = p^2 + q^2 + 2pq - 2p - 2q
⇒ pq = p + q
⇒ p = q/q-1
Given that p,q, r are integers and as q and (q-1) are co-primes, we have,
For p to be an integer, q - 1 = ±1
⇒ q = 0 or 2
when, q = 0 ⇒ p = 0/0-1 = 0
∴ p = 0
using (1), r = p + q - 1 = 0 + 0 - 1 = -1
∴ r = -1.
∴ p + q + r = 0 + 0 + (-1) = -1
when, q = 2 ⇒ p = 2/2-1 = 2
∴ p = 2
using (1), r = p + q - 1 = 2 + 2 - 1 = 3
∴ r = 3.
∴ p + q + r = 2 + 2 + 3 = 7
∴ Sum of all positive values of (p + q - r) = (7 - 1) = 6
Given : Integers p,q,r satisfy p+q-r=-1 and p^2+q^2-r^2=-1
To find : sum of all positive values of p+q+r
Solution:
p+q-r=-1
=> p + q = r - 1
Squaring both sides
=> p² + q² + 2pq = r² + 1 - 2r
p²+q²-r²=-1
=> p²+q² = r² -1
Substituting this value
=> r² -1 + 2pq = r² + 1 - 2r
=> 2pq = 2 - 2r
=> pq = 1 - r
=> pq = - (r - 1)
=> pq = - (p + q)
=> p + q + pq = 0
=> p = -q/(q + 1) & q = -p/(p + 1)
p , q & r are integer so q + 1 or p + 1 = ± 1
only possible Values
p = q = 0 or p = q = -2
=>pq = 0 or pq = 4
pq = 1 - r => r = 1 - pq
p = q = 0 => r = 1 - 0 = 1
=> p + q + r = 1
p = q = -2 => r = 1 - 4 = - 3
p + q + r = -2 - 3 = - 5 ( negative Value)
positive values of p+q+r. = 1 only
sum of all positive values of p+q+r. = 1
Learn more:
If the ordered triplets of real numbers (x, y, z) satisfy √x-y+z=√x ...
https://brainly.in/question/18708970
if x+y+z=0 then the square of the value of (x+y)^2/xy+(y+z)^2/yz+(z+x)
https://brainly.in/question/12858114