suppose a quadratic polynomial p(x) = ax^2 + bx + c has positive coefficients a,b,c in AP in that order. if p(x) = 0 has integer roots p and q, then p + q +pq =
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P(x)=ax∧2+bx+c
and roots are p and q
then
roots=(-b+√(b∧2-4ac) )/2a , (-b-√(b∧2-4ac) )/2a
let p = (-b+√(b∧2-4ac) )/2a , q = (-b-√(b∧2-4ac) )/2a
p+q+pq=(-b+√(b∧2-4ac) )/2a + (-b-√(b∧2-4ac) )/2a +(-b+√(b∧2-4ac) )/2a * (-b-√(b∧2-4ac) )/2a
= -2b/2a+ ( (-b)∧2 - (√(b∧2-4ac)∧2 )/2a
=-b/a+(b∧2-b∧2+4ac)/4a∧2
=b/a+4ac/4a∧2
=b/a+c/a
=(b+c)/a
and roots are p and q
then
roots=(-b+√(b∧2-4ac) )/2a , (-b-√(b∧2-4ac) )/2a
let p = (-b+√(b∧2-4ac) )/2a , q = (-b-√(b∧2-4ac) )/2a
p+q+pq=(-b+√(b∧2-4ac) )/2a + (-b-√(b∧2-4ac) )/2a +(-b+√(b∧2-4ac) )/2a * (-b-√(b∧2-4ac) )/2a
= -2b/2a+ ( (-b)∧2 - (√(b∧2-4ac)∧2 )/2a
=-b/a+(b∧2-b∧2+4ac)/4a∧2
=b/a+4ac/4a∧2
=b/a+c/a
=(b+c)/a
anurajpikeovx7wt:
sorry it would be (c-b)/a
last three step i forgor - sign before b
Actually the question had asked for a value..
It was a KVPY question..
But thanx for your help...
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