Question

Find the zeroes of the polynomial (x-2)² + 4.​

Answer 1

Answer:

b² -4ac = -16 < 0, No real roots.

Explanation:

Let p(x) = (x-2)² + 4

To find: The zeroes of p(x)

Proof:

= p(x) = (x-2)² + 4

Let p(x) = 0

⇒ p(x) = (x-2)² + 4 = 0

⇒ x² - 2(x)(2) + 2² + 4 = 0                            [ (a-b)² = a² -2ab +b² ]  

⇒ x² - 4x + 4 + 4 = 0

⇒ x² - 4x +8 = 0

By discriminant formula,

• b = -4
• a = 1
• c = 8

D = b² -4ac

= (-4)² - 4(1)(8)

= 16 - 32

= -16

As b² -4ac = -16 < 0

∴This equation has no real roots.

Hence, proved.

Hope you got that.

Thank You