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 > $ \frac { 5 }{ 3 }$
- a < $\frac { 5 }{ 3 }$
- a = $\frac { 5 }{ 3 }$
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a > 2
For real roots, the discriminant D = b^2 - 4ac >= 0. Here, D = [2(3a+5)]^2 - 4(1)(2(9a^2+25)) = 4(9a^2 + 30a + 25) - 8(9a^2 + 25) = 36a^2 + 120a + 100 - 72a^2 - 200 = -36a^2 + 120a - 100. Setting -4(9a^2 - 30a + 25) = -4(3a-5)^2 >= 0. Since a square is always non-negative, this is only >= 0 when 3a-5 = 0, so a = 5/3.
For a quadratic equation to have real roots, its discriminant (D = b^2 - 4ac) must be greater than or equal to zero. Here, a = 1, b = 2(3a + 5), and c = 2(9a^2 + 25), so applying the discriminant condition gives [2(3a + 5)]^2 - 4(1)(18a^2 + 50) >= 0. Expanding this yields 36a^2 + 120a + 100 - 72a^2 - 200 >= 0, which simplifies to the quadratic inequality -36a^2 + 120a - 100 >= 0, or equivalently 9a^2 - 30a + 25 <= 0. Factoring this perfect square gives (3a - 5)^2 <= 0; since the square of any real number cannot be negative, it must equal zero, meaning 3a - 5 = 0 and a = 5/3.