Question

If a,b,c >0 and $a=2b+3c$, then the roots of the equation $a{x}^2+bx+c=0$ are real if

• A. $\mid \frac{a}{c}-11\mid \ge 4\sqrt{7}$
• B. $\mid \frac{c}{a}-11\mid >3\sqrt{7}$
• C. $\mid \frac{b}{c}-4\mid \ge 4\sqrt{7}$
• D. $\mid \frac{c}{b}-4\mid \ge 2\sqrt{7}$

Answer 1

The correct option is A $\mid \frac{a}{c}-11\mid \ge 4\sqrt{7}$

Given $a,b,c>0$

$a=2b+3c$-Equation 2

$f\left(x\right)=a{x}^2+bx+c=0$

For real roots, $D\ge 0$

$\Rightarrow D\equiv b^2-4ac\ge 0$

$\equiv (\frac{a-3c}{2}{)}^2-4ac\ge 0$

$\Rightarrow a^2-6ca+9c^2-16ac\ge 0$

$\Rightarrow a^2-22ac+9c^2\ge 0$

$\Rightarrow (\frac{a}{c}{)}^2-22\frac{a}{c}+9\ge 0$

$\Rightarrow (\frac{a}{c}{)}^2-\frac{22}{9}(\frac{a}{c})+9+(11{)}^2-(11{)}^2\ge 0$

$\Rightarrow (\frac{a}{c}-11{)}^2-112\ge 0$

$\Rightarrow \left[\right(\frac{a}{c}-11{)}^2]-112\ge 0$

$\Rightarrow \left[\right(\frac{a}{c}-11{)}^2]\ge 112$

Taking square root, we get $\mid (\frac{a}{c}-11)\mid \ge 4\sqrt{7}$