Finding Roots: Solving 5x^2 + 7x + b + a = 0 Under Inequality Constraint

Question

Given the following equation, given also 4920a+b \frac{49}{20}\ge a+b

5x2+7x+b+a=0 5x^2+7x+b+a=0

Video Solution

Solution Steps

00:00 Find X
00:07 Identify the coefficients
00:20 Use the root formula to find possible solutions
00:26 Substitute appropriate values and solve to find solutions
00:39 Calculate the square and products
00:58 Range of the sum of unknowns according to the given data
01:12 If we use this range, the root must be greater than/equal to 0
01:21 Therefore the equation is logical
01:24 And this is the solution to the question

Step-by-Step Solution

To solve the quadratic equation 5x2+7x+b+a=0 5x^2 + 7x + b + a = 0 , we will use the quadratic formula. The equation is in the form ax2+bx+c=0 ax^2 + bx + c = 0 , where a=5 a = 5 , b=7 b = 7 , and c=a+b c = a + b .

The quadratic formula is given by:

x=b±b24ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substituting the known values, we have:

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

First, calculate the discriminant:

b24ac=7245(a+b) b^2 - 4ac = 7^2 - 4 \cdot 5 \cdot (a + b)

=4920(a+b) = 49 - 20(a + b)

Substituting into the quadratic formula, we get:

x=7±4920(a+b)10 x = \frac{-7 \pm \sqrt{49 - 20(a + b)}}{10}

We also have a given condition: 4920a+b \frac{49}{20} \ge a + b .

Since the discriminant 4920(a+b) 49 - 20(a + b) needs to be non-negative for real solutions, the given condition ensures that the discriminant is non-negative, as a+b4920 a + b \le \frac{49}{20} means 4920(a+b)0 49 - 20(a + b) \ge 0 .

Therefore, the solution to the problem is:

7±4920(a+b)10 \frac{-7 \pm \sqrt{49 - 20(a + b)}}{10}

This corresponds to choice 3 in the given options.

Answer

7±4920(a+b)10 \frac{-7\pm\sqrt{49-20(a+b)}}{10}