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Vedmedyk [2.9K]
3 years ago
11

flvs hope If $x^2+bx+16$ has at least one real root, find all possible values of $b$. Express your answer in interval notation.

Enter your answer Focus Quadratic Inequalities 18 Week 5 Quests © 2018 Art of Problem Solving Terms Privacy Contact Us About Us ap
Mathematics
1 answer:
Galina-37 [17]3 years ago
5 0
If x^2+bx+16 has at least one real root, then the equation x^2+bx+16=0 has at least one solution. The discriminant of a quadratic equation is b^2-4ac and it determines the nature of the roots. If the discriminant is zero, there is exactly one distinct real root. If the discriminant is positive, there are exactly two roots. The discriminant of <span>x^2+bx+16=0 is b^2-4(1)(16). The inequality here gives the values of b where the discriminant will be positive or zero:
                                           b^2-4(1)(16) ≥ 0
                                                  </span><span>b^2-64 ≥ 0
                                             (b+8)(b-8) </span><span>≥ 0
The answer is that all possible values of b are in the interval (-inf, -8]∪[8,inf) because those are the intervals where </span>(b+8)(b-8) is positive.
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