Question

If the equation $c{x}^2+bx-2a=0$ has no real roots and $a<\frac{b+c}{2}$, then choose the wrong option.

• A. $ac<0$
• B. $a<0$
• C. $\frac{c-b}{2}>a$
• D. $\frac{c+2b}{8}<a$

Answer 1

Answer: (D)

$\frac{c+2b}{8}<a$

The equation $c{x}^2+bx-2a=0$ has non-real roots .
Hence, the expression $c{x}^2+bx-2a$ must either be positive or negative for all real values of $x$ .
We have,
$a<\frac{b+c}{2}$
$\Rightarrow b+c-2a>0$
$\Rightarrow c(1{)}^2+b(1)-2a>0$
Hence, the expression $c{x}^2+bx-2a$ is positive for $x=1$ .
Hence, it must be positive for all values of $x$. This also implies that the coefficient of $x^2$ is positive .
Hence, $c>0$ .
Now, $c\left(0\right)+b\left(0\right)-2a>0$
$\Rightarrow a<0$
Hence, $ac<0$
Similarly,
$c(-1{)}^2+b(-1)-2a>0$
$c-b>2a$
$\frac{c-b}{2}>a$
Hence, all the options are correct .
Similarly, consider $c{\left(\frac{1}{2}\right)}^2+b\left(\frac{1}{2}\right)-2a>0$
$\Rightarrow c+2b-8a>0$
$\Rightarrow \frac{c+2b}{8}>a$
So, option D is not correct.