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Let a,b be arbitary real numbers. Find the smallest natural number `b` for which the equation x^2+2 (a+b)x+(a-b+8)=0 has unequal real roots for all a belongs to real numbers.

Let a,b be arbitary real numbers. Find the smallest natural number `b` for which the equation x^2+2 (a+b)x+(a-b+8)=0 has unequal real roots for all a belongs to real numbers.

Grade:11

2 Answers

Ritesh Khatri
76 Points
6 years ago
(a-b+8) > 0If D 2 – 4
a2 + b2 + 2ab – a – b – 8 >  0
a2 + (2b – 1)a + b2 – 8 >  0
it is given a belongs to real numbers and abobe quadratic is always greater than 0 hence polynomial a2 + (2b – 1)a + b2 – 8 can not have real roots 
hence its D
(2b – 1)2 – 4(b2 – 8) 
 
- 4b + 1 + 24 
b > 25/4
 
smallest naturan number b = 7
 
Ritesh Khatri
76 Points
6 years ago
Initial line is for unequal roots discriminant of given equation must be greater than 0, D > 0 
(a-b+8) > 0(2 (a+b))^2 – 4
 

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