Question

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

Answer 1

Answer:

Option 4 is correct.

Step-by-step explanation:

To find: Polynomial whose solution are not real numbers.

Given Polynomials are Quadratic Polynomial.

So, we can check if solution of quadratic polynomial by find & checking value of discriminant.

Standard form of Quadratic polynomial is given by

ax² + bx + c

then Discriminant, D = b² - 4ac

If, D > 0 ⇒ Solutions are distinct real numbers

if, D = 0 ⇒ Solutions are equal real numbers

if, D < 0 ⇒ Solutions are not real numbers (They are complex conjugates)

Option A:

By comparing with standard form

a = 1 , b = -6 , c = 3

D = (-6)² - 4 × 1 × 3 = 36 - 12 = 24 > 0

Thus, Solutions are Real numbers.

Option B:

By comparing with standard form

a = 1 , b = 4 , c = 3

D = (4)² - 4 × 1 × 3 = 16 - 12 = 4 > 0

Thus, Solutions are Real numbers.

Option C:

By comparing with standard form

a = -1 , b = -9 , c =-10

D = (-9)² - 4 × (-1) × (-10) = 81 - 40 = 41 > 0

Thus, Solutions are Real numbers.

Option D:

By comparing with standard form

a = 1 , b = 2 , c = 3

D = (2)² - 4 × 1 × 3 = 4 - 12 = -8 < 0

Thus, Solutions are not Real numbers.

Therefore, Option 4 is correct.

Answer 2

I maybe totally wrong but i want to say either A or C. A nonreal number is any number that when multiplied by itseld produces a negative number.