the roots of the equation x^2+(p+2)x+2p=0 …
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Posted by Abhay Singh 7 years, 10 months ago
- 1 answers
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Dharmendra Kumar 7 years, 10 months ago
x2+(p+2)x+2p=0
On comparing with ax2+bx+c=0 we get
a=1 b=p+2 c=2p
Now since roots of the given quadratic equation are distinct integer.
Therefore, Discriminent D=b2-4ac > 0
(p+2)2-4×1×2p > 0
p2+2×p×2+22 -8p > 0
p2+4p+4-8p > 0
p2 -4p +4 > 0
p2-2p-2p+4 > 0
p(p-2)-2(p-2) > 0
(p-2)(p-2) > 0
(p-2)2 > 0
p-2 > 0
p > 2
So roots of the given quadratic eq. is distinct integer if p>2.
2Thank You