If a, b, c are +Ve and a = 2b+3c, then roots of the equation ax²+bx+c = 0 are real for

The given equation is $a{x}^2+bx+c=0$

Since the roots are real $b^2-4ac\ge 0$

given $a=2b+3c$

Elimiante $b$, we get

$\begin{array}{rcl}{\left(\frac{a-3c}{2}\right)}^2-4ac& \ge & 0\Rightarrow a^2+9c^2-6ac-16ac\ge 0\\ & \Rightarrow & {\left(\frac{a}{c}\right)}^2-22\left(\frac{a}{c}\right)+9\ge 0\\ & \Rightarrow & {\left(\frac{a}{c}-11\right)}^2\ge 112\\ & \Rightarrow & \left|\frac{a}{c}-11\right|\ge 4\sqrt{7}\\ & & \end{array}$

Eliminate $a$ we get

$\begin{array}{rcl}{b}^2-4ac& \ge & 0\\ b^2-4\left(2b+3c\right)c& \ge & 0\\ b^2-8bc+12c^2& \ge & 0\\ {\left(\frac{b}{c}\right)}^2-8\frac{b}{c}-12& \ge & 0\\ {\left(\frac{b}{c}-4\right)}^2& \ge & 28\\ \left|\frac{b}{c}-4\right|& \ge & 2\sqrt{7}\end{array}$