Question

Find $a$ so that roots of $x^2+2(3a+5)x+2(9a^2+25)=0$ are real.

Answer 1

For roots to be real,

$D\ge$0

So , $4(3a+5{)}^2-4\times 1\times 2(9a^2+25)\ge 0$

$\to 9a^2+25+30a-18a^2-50\ge 0$

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

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

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

But square of a real number can't be less than 0

So only equality can hold

Thus a=$\frac{5}{3}$