Question

If the roots of the equation $x^2+2(3a+5)x+2(9a^2+25)=0$ are real, then what will be the value of 'a'?

• A. $a>\frac{5}{3}$
• B. $a<\frac{5}{3}$
• C. $a=\frac{5}{3}$
• D. $a>2$

Answer 1

The correct option is C $a=\frac{5}{3}$

Roots of the given equation are real, If $D=b^2-4ac\ge 0$ i.e., if $4(3a+5{)}^2-8(9a^2+25)\ge 0$ i.e., if $(3a+5{)}^2-2(9a^2+25)\ge 0$ i.e., if $9a^2+30a+25-18a^2-50\ge 0$ i.e., if $9a^2-30a+25\le 0$ i.e., if $(3a-5{)}^2\le 0$ or if $3a-5=0$ or if $a=5/3$