Question

If $2a+3b+6c=0,$ then the equation $a{x}^2+bx+c=0$ has at least one real root in

• A. $(0,1)$
• B. $(0,\frac{1}{2})$
• C. $(\frac{1}{4},\frac{1}{2})$
• D. $(-1,1)$

Answer 1

The correct option is A $(0,1)$

$2a+3b+6c=0$

$⟹b=-\frac{2(a+3c)}{3}$

$a{x}^2+bx+c=0$

$x=\frac{-b+-\sqrt{(b^2-4ac)}}{2a}$

=$\frac{\frac{2(a+3c)}{3}\pm \sqrt{[\frac{4}{9}\times (a+3c{)}^2-4ac]}}{2a}$

= $\frac{a+3c\pm \sqrt{[a^2+9c^2-3ac]}}{\left(3a\right)}$

EITHER, $(a-3c{)}^2\le [a^2+9c^2-3ac]\le (a+3c{)}^2$ OR, $(a+3c{)}^2\le (a^2+9c^2-3ac)\le (a-3c{)}^2$

The boundaries are for one root are :

$\frac{[a+3c+(a-3c\left)\right]}{\left(3a\right)}=\frac{2}{3}$ and $\frac{[a+3c+(a+3c\left)\right]}{\left(3a\right)}=2\frac{(a+3c)}{\left(3a\right)}=-b/$a

the boundaries or limits for the other root are:

$\frac{[a+3c-(a-3c\left)\right]}{\left(3a\right)}=\frac{2c}{a}$ and $\frac{[a+3c-(a+3c\left)\right]}{\left(3a\right)}=\frac{2}{3}$

So one root is between $\frac{2}{3}$ and $\frac{-b}{a}$

the other is between $\frac{2}{3}$ and $\frac{2c}{a}$

Therefore lies in $(0,1)$