Question

If the roots of the equation $x^2+2(3a+5)x+2(9a^2+25)=0$ are real, then find $a$.

• A. $\frac{5}{3}$
• B. $\frac{7}{3}$
• C. $\frac{2}{3}$
• D. None of these

Answer 1

The correct option is A $\frac{5}{3}$

As roots of equation are real

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

$\Rightarrow \left[2\right(3a+5){]}^2-4\times 1\times (18a^2+50)\ge 0$

$\Rightarrow 4(9a^2+25+30a)-72a^2-200\ge 0$

$\Rightarrow 36a^2+100+120a-72a^2-200\ge 0$

$\Rightarrow 36a^2+100-120a\le 0$

$\Rightarrow (6a-10{)}^2\le 0\Rightarrow a\le \frac{5}{3}$