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Which of the following polynomials has solutions that are not real numbers?

A)x^2-6x+3
B)x^2+4x+3
C)-x^2-9x-10
D)x^2+2x+3

User Nereida
by
7.8k points

1 Answer

4 votes

Answer:

D)
x^(2) + 2x + 3.

Explanation:

All four polynomials are quadratic, meaning that the highest power of the unknown
x in the equation is two.

The sign of the quadratic discriminant,
\Delta, is a way to tell if a quadratic polynomial comes with non-real solutions.

There are three cases:


  • \Delta > 0. The quadratic discriminant is positive. There are two real solutions and no non-real solution. The two solutions are different from each other.

  • \Delta = 0. The quadratic discriminant is zero. There is one real solution and no non-real solution.

  • \Delta <0. The quadratic discriminant is negative. There is no real solution and two non-real solutions.

How to find the quadratic discriminant?

If the equation is in this form:


a \; x^(2) + b\;x + c = 0,

where a, b, and c are real numbers (a.k.a. "constants.")

Quadratic discriminant:


\Delta = {b^(2) - 4\;a\cdot c}.

Polynomial in A:


x^(2) - 6x + 3 = 0.


  • a = 1.

  • b = -6.

  • c = 3.


\Delta = b^(2) - 4 \;a\cdot c = (-6)^(2) - 4* 1* 3 = 36 - 12 = 24.


\Delta > 0. There will be no non-real solutions and two distinct real solutions.

Try the steps above for the polynomial in B, C, and D.

  • B):
    \Delta = 4. Two distinct real solutions. No non-real solution.
  • C):
    \Delta= 41. Two distinct real solutions. No non-real solution.
  • D):
    \Delta = -8. No real solution. Two distinct non-real solutions.
User Ramanman
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