Question

If $c>0$ and $4a+c<2b$, then $a{x}^2-bx+c=0$ has a root in the interval

• A. $(0,2)$
• B. $(2,4)$
• C. $(-2,0)$
• D. $(4,9)$

Answer 1

Answer: (A)

$(0,2)$

Let $f\left(x\right)=a{x}^2-bx+c$

Given $4a+c<2b,c>0$

$4a-2b+c<0$

$f\left(0\right)=c$

$f\left(0\right)>0$

$f\left(2\right)=4a-2b+c$

$f\left(2\right)<0$

Because $f\left(x\right)$ is continuous and $f\left(0\right)>0,f\left(2\right)<0$

So, $f\left(x\right)$ has a root in interval $(0,2)$