First slide
Theory of expressions
Question

Given that ax2+bx+c=0 has no real roots and a+b+c<0 then

Moderate
Solution

Since the equation ax2+bx+c=0 has no real roots. Therefore, the curve y=ax2+bx+c does not intersect with x-axis. Consequently ϕ(x)=ax2+bx+c c has same sign for all values of x. It is given that 

a+b+c<0ϕ(1)=a+b+c<0

ϕ(x)<0 for all xϕ(0)<0c<0

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