Question

The coefficient of the quadratic equation ${\mathrm{ax}}^2+(a+d)x+(a+2d)=0$ are consecutive terms of a positively valued, increasing arithmetic sequence. Determine the least integral value of $\frac{d}{a}$ such that the equation has real solutions.

Answer 1

Answer: 7

$a{x}^2+(a+d)x+(a+2d)=0$
$a,a+d,a+2d$ are in $\uparrow$ A.P. $(d>0)$ (note that positively valued terms $\Rightarrow a>0$ )
for real roots $D\ge 0$
$\Rightarrow (a+d{)}^2-4a(a+2d)\ge 0$
$\Rightarrow a^2+d^2+2ad-4a^2-8ad\ge 0$
or $d^2-3a^2-6ad\ge 0\text{ or }{t}^2-6t-3\ge 0\text{ where }t=\frac{d}{a}\text{ and }t>0$
$(d-3a{)}^2-12a^2\ge 0$
$(d-3a-\sqrt{12}a)(d-3a+\sqrt{12}a)\ge 0$

$\left[\frac{d}{a}-(3+2\sqrt{3})\right]\left[\frac{d}{a}-(3-2\sqrt{3})\right]\ge 0$
$\therefore$ d $=3+2\sqrt{3}\ge 6\Rightarrow$ least integral value $=7$