See Exercise 47.
(b)Which equation has two nonreal complex soluti...

Question

See Exercise 47.
(b)Which equation has two nonreal complex solutions?

Accepted Answer

Which of the following quadratic equations has two non-real complex roots. We noticed here we have four different quadratic equations. Let's check every one of these to determine what the roots are. Our first one. A six X -11 multiplied by six X -11 is equals to zero. We actually have a duplicate here because we have six X -11 twice. So six x -11 Squared is equals to zero. Solving for X, we would end up getting one real route. Let's check answer B for B you have nine minus eight X squared is equals to zero. Let's first say the square of both sets square -8 X is equals to zero. If we were to keep solving this, we would then figure out we have one real route. So let's check answer C now see if five X minus eight squared is equals to negative 16, take square to both sides get five X minus eight S equals squared of 16, four, I, if we were to keep going, we'd have an imaginary number at the end or a complex number By adding eight and dividing by five. In that case, we should have two complex routes. Should also note we take the spirit of -16. So we will actually have a plus or -4, which accounts for a second group for D, we'll check D as well. We have 10 x plus nine square as equals 15, a square. On both sides, you will get 10 X plus nine as equals to squared at 15. If we were to keep solving this, we would get two real roots. There's only one answer then with two complex roots which corresponds to answer C OK. I hope to help you solve the problem. Thanks for watching. Goodbye.

See Exercise 47. (b)Which equation has two nonreal complex solutions?

Key Concepts

Complex Numbers

Quadratic Equations

Discriminant

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