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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.
(a-b+8) > 0If D 2 – 4a2 + b2 + 2ab – a – b – 8 > 0a2 + (2b – 1)a + b2 – 8 > 0it 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
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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