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