Math, asked by adfj, 1 year ago

find the zeros of the polynomial (x-2)square + 4

Answers

Answered by Anonymous
34
Hey Mate

your answer is ---

we have ,

(x-2)^2 +4

= x^2 +4 - 4x + 4

= x^2 - 4x + 8

for zeroes

x^2 - 4x + 8 = 0

such that , a = 1 , b = -4 & c = 8

now, by quadratic formula for value of x

x = ( -4 ± √-4^2+4×8×1) / 2×1

=> x = ( -4 ± √48 ) / 2

=> x = (-4 ± 4√3) / 2

=> x = (-4 + 4√3 ) / 2 or. (-4 - 4√3) /3

=> x = -2(1-√3) or -2(1+√3)




【 HOPE IT HELP YOU 】

adfj: thnx for your ans
Anonymous: welcome
Answered by harshi229
4

Answer:

Step-by-step explanation:

The number of zeros the given polynomial (x - 2)² + 4 has = 2

Step-by-step explanation:

The polynomial is given to be : (x - 2)² + 4

To find the number of zeros of the polynomial we first simplify the given polynomial

⇒ (x - 2)² + 4

⇒ x² + 4 - 4x + 4

⇒ x² - 4x + 8

So, It can be seen that the degree of the given polynomial is 2

So, It can have 2 zeros

Hence, The number of zeros the given polynomial (x - 2)² + 4 has = 2

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