if zeroes of the polynomial ax^2+bx+c are p and q then find the value of p/q+q/p
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Answered by
2
Answer:
b²/ac - 2
Step-by-step explanation:
The sum of the roots is equal to -b/a. The product of the roots is equal to c/a.
[ To see why, note that ax²+bx+c = a(x-p)(x-q) since p and q are root, then equate coefficients. ]
So...
p+q = -b/a
pq = c/a
From here, we get
p²+q² = (p+q)² - 2pq = b²/a² - 2c/a
Then
p/q + q/p
= (p²+q²) / pq
= ( b²/a² - 2c/a ) / (c/a)
= b²/ac - 2
Anonymous:
You're welcome. Glad to have helped.
Plz mark brainliest
thx!
I didn't get it
Same approach. Just need to get expressions p+q = -b/a and pq = c/a right.
sorry for that because it was in my 9th std syllabus now it had been 5 years
sorry for the small mistake
No problem. :D Notice it is easy to work out the formulas when you need them, so you don't need to remember them.
ok
ok
Answered by
0
p+q=a/c
pa=-b/c
p/q+q/p=(p^2+q^2)/pq=[(p+q)^2-2pq]/pq
[(p+q)^2-2pq]/pq=[(a/c)^2-2(-b/c)]/(-b/c)=(a^2/c^2+2b/c)/(-b/c)=(a^2+2bc/c^2)/(-b/c)=by canceling c we get
-(a^2+2bc)/bc
you can substitute by dividing -(a^2/bc+2)
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