Let \( a \) and \( b \) be the roots of the quadratic equation \( 2x^2 + 5x + 30 = 0 \).
By Vieta's formulas, the sum of the roots is
\[ a + b = -\frac{5}{2} \]
and the product of the roots is
\[ ab = \frac{30}{2} = 15. \]
We want to find the value of \( a^2 + b^2 \). We know that
\[ (a + b)^2 = a^2 + 2ab + b^2. \]
Then we have
\[ a^2 + b^2 = (a + b)^2 - 2ab. \]
Substituting the values we found earlier, we have
\[ a^2 + b^2 = \left( -\frac{5}{2} \right)^2 - 2(15) = \frac{25}{4} - 30 = \frac{25}{4} - \frac{120}{4} = \frac{25 - 120}{4} = \frac{-95}{4}. \]
Therefore, \( a^2 + b^2 = -\frac{95}{4} \).
Final Answer: The final answer is \( \boxed{-\frac{95}{4}} \)
| List-I | List-II | ||
|---|---|---|---|
| (A) | $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ | (I) | $[\frac{1}{3} , 1 ]$ |
| (B) | $(x)=|[x]|,x \in [R$ | (II) | Z |
| (C) | $h(x) = |x - [x]| , x \in [R$ | (III) | W |
| (D) | $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ | (IV) | [0, 1) |
| (V) | { -1, 1} | ||
| List I | List II | ||
|---|---|---|---|
| (A) | $\lambda=8, \mu \neq 15$ | 1. | Infinitely many solutions |
| (B) | $\lambda \neq 8, \mu \in R$ | 2. | No solution |
| (C) | $\lambda=8, \mu=15$ | 3. | Unique solution |