Question

Show that the roots of the equation $x^2+2(3a+5)x+2(9a^2+25)=0$ are complex unless $a=\frac{5}{3}$

Answer 1

For Real roots $\Rightarrow D\ge 0$

$\Rightarrow \left(2\right(3a+5){)}^2-4(1\left)\right(2(a_a^2+25))\ge 0$

$\Rightarrow 4(a_a^2+25+30a)-4(18a^2+50)\ge 0$

$\Rightarrow -9a^2+30a-25\ge 0$

$\Rightarrow 9a^2-30a+25\le 0$

$\Rightarrow (3a-5{)}^2\le 0$

$\Rightarrow (3a-5{)}^2=0$

$\Rightarrow$ Only possible if $a=\frac{5}{3}$