If the roots of the equation ${ x }^{ 2 }+2(3a+5)x+2(9{ a }^{ 2 }+25)=0$ are real, then find $a$.
- $\cfrac{5}{3}$
- $\cfrac{7}{3}$
- $\cfrac{2}{3}$
-
None of these
For real roots, discriminant D >= 0. D = (2(3a+5))^2 - 4(1)(2(9a^2+25)) >= 0. 4(9a^2 + 30a + 25) - 8(9a^2 + 25) >= 0. 36a^2 + 120a + 100 - 72a^2 - 200 >= 0. -36a^2 + 120a - 100 >= 0. Dividing by -4: 9a^2 - 30a + 25 <= 0. (3a-5)^2 <= 0. This only holds if 3a-5 = 0, so a = 5/3.
For the roots to be real, the discriminant (D = b^2 - 4ac) must be greater than or equal to zero. Substituting the values gives [2(3a + 5)]^2 - 4 * 1 * 2(9a^2 + 25) >= 0, which simplifies to 4(9a^2 + 30a + 25) - 8(9a^2 + 25) >= 0. Expanding and combining like terms results in -36a^2 + 120a - 100 >= 0, and dividing by -4 gives 9a^2 - 30a + 25 <= 0. This factors into the perfect square (3a - 5)^2 <= 0, meaning the only possible solution is a = 5/3.