Question

Find the zero of the polynomial (x-2)^+4

Answer 1

Answer:

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)

Answer 2

Hlo.. ans:⤵

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)

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