Let a,b be arbitary real numbers. Find the smalles

Question

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

Dinesh Nayak · Accepted Answer

we have , x2+2(a+b)+(a-b+8)=0 => x2+3a+b+8=0 for the equation to have unequal real roots then it should have the discriminant b2-4ac>0 so we get, b2-4ac>0=> 0-4(1)(3a+b+8)>0=> -12a-4b-32>0 on solving the in equation we get the range for ‘b’ as b ( - , -3a-8) so according to the range if a = -(8/3)then b=0so the smallest natural number for b is 0

Algebra> Let a,b be arbitary real numbers. Find th...

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

Deep shikha , 7 Years ago

Dinesh Nayak

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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)+(a-b+8)=0 has unequal real roots for all a belongs to real numbers

Deep shikha , 7 Years ago

Dinesh Nayak

LIVE ONLINE CLASSES

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