Quadratic Equations

Question

Consider the quadratic equation x^2 + bx + 10 = 0, where b is a constant. Determine possible values of b so that this question does not have a solution.

This question is very confusing. The answer says -2√10 < x < 2√10. I would like help on how to get that answer as I've got no clue. Thanks!

Answer 1

Quadratic equations ALWAYS have 2 solutions.

ASSuming you mean, "does not have a REAL solution" then that happens when the

discriminant is NEGATIVE.

In that case, b^2 - 4(1)(10) < 0

b^2 - 40 < 0

(b - sqrt(40)) (b + sqrt(40)) < 0

THis happens when the factors are opposite signs

So then if the factors have opposite signs, then the following condition MUST hold:

So then - sqrt(40) < b < sqrt(40)

BUT sqrt(40) = sqrt( 4*10) = sqrt(4) * sqrt(10) = 2 * sqrt(10)

THis leads to the answer you have shown in your post.