Finding Roots: Solving 5x² + 7x + b + a = 0 Under Inequality Constraint

Given the following equation, given also $\frac{49}{20}\ge a+b$

$5x^2+7x+b+a=0$

To solve the quadratic equation $5x^2 + 7x + b + a = 0$, we will use the quadratic formula. The equation is in the form $ax^2 + bx + c = 0$, where $a = 5$, $b = 7$, and $c = a + b$.

The quadratic formula is given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting the known values, we have:

$a = 5$, $b = 7$, $c = a + b$

First, calculate the discriminant:

$b^2 - 4ac = 7^2 - 4 \cdot 5 \cdot (a + b)$

$= 49 - 20(a + b)$

Substituting into the quadratic formula, we get:

$x = \frac{-7 \pm \sqrt{49 - 20(a + b)}}{10}$

We also have a given condition: $\frac{49}{20} \ge a + b$.

Since the discriminant $49 - 20(a + b)$ needs to be non-negative for real solutions, the given condition ensures that the discriminant is non-negative, as $a + b \le \frac{49}{20}$ means $49 - 20(a + b) \ge 0$.

Therefore, the solution to the problem is:

$\frac{-7 \pm \sqrt{49 - 20(a + b)}}{10}$

This corresponds to choice 3 in the given options.