Question

2. Consider the equation a x + bx + c = 0. Assume that the discriminant is zero and that a is positive. 2 What are the solutions of the inequality a x + bx + c < 0?

Answer 1

Answer:

The equation ax²+bx+c has no solutions

Step-by-step explanation:

I am assuming that the equation is ax²+bx+c = 0. Lets call r₁ and r₂ the roots, then

$ax^2+bx+c = a(x-r_1)(x-r_2)$

The roots r₁ and r₂ can be computed with the quadratic formula

$r_1, r_2 = \frac{-b^+_-\sqrt{b^2-4ac}}{2a}$

The discriminant is zero, therefore b²-4ac = 0, and as a result

$r_1, r_2 = \frac{-b}{2a}$

Since both roots are equal, then we can call them r, with r equal to -b/2a. Replacing this in the factorization gives us as a result

$ax^2+bx+c = a(x-r)²$

Note that a is positive and (x-r)² is zero at least, this means that the product is equal or greater than zero for every value of x. We conclude that the equation a x² + bx + c < 0 has no solutions.