If c<0 and ax2+bx+c=0 does not have any real roots then prove that 9a+3b+c<0.
Prove the given expression
The given condition is c<0, f(x)=ax2+bx+c
At x=0,f(x)is
f(0)=c∴f(0)<0[∵c<0]
Also, f(x) does not have any real roots
∴ We can say for all x∈ℝ,f(x)<0
Let x=3
f(3)=9a+3b+c∵f(x)<0∀x∈ℝ∴f(3)<0⇒9a+3b+c<0
Hence proved, 9a+3b+c<0.
The equation ax2+bx+c = 0 does not have real roots and c < 0. Which of these is true?
if a,b,c are real numbers such that ac is not equal to 0,then show that at least one of the equations ax2 + bx + c = 0 and -ax2 + bx + c = 0 has real roots.