Question

if ax^2 + bx + 6=0 doesn't have 2 distinct real roots, then find the least value of 3a + b

Answer 1

Answer:

Least value of 3a+b is -2

Explanation:

Given x²+bx+6= 0 doesn't have two distinct real roots,

so b²=4ac

b²=4a×6= 24a

b²=24a

3a= b²\b

3a+b=( b²\8)+b = (b²+8b)\8 = (b²+2.4b+16-16)\8b= (b + 4)²\8b

minimum value of (b + 4)²=0

So minimum value of {(b + 4)²-16}/8b = 0-16\8 = -2

Here applied rule is

if any second degree quadrilateral equation doesn't have 2 distinct roots then b²=4ac