(x-5)^2+3(x-5)+9=0 what is solution
1 answer:
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Answer:
a) zeros of the function are x = 1 and, x = -1
b) zeros of the function are x = 2 and, x = 4
c) zeros of the function are x =
d) zeros of the function are x = and, x = 17
Step-by-step explanation:
Zeros of the function are the values of the variable that will lead to the result of the equation being zero.
Thus,
a) f(x)= (x +1)(x − 1)(x² +1)
now,
for the (x +1)(x − 1)(x² +1) = 0
the condition that must be followed is
(x +1) = 0 ..........(1)
or
(x − 1) = 0 ..........(2)
or
(x² +1) = 0 ...........(3)
considering the equation 1, we have
(x +1) = 0
or
x = -1
for
(x − 1) = 0
x = 1
and,
for (x² +1) = 0
or
x² = -1
or
x = √(-1) (neglected as it is a imaginary root)
Thus,
zeros of the function are x = 1 and, x = -1
b) g(x) = (x − 4)³(x − 2)⁸
now,
for the (x − 4)³(x − 2)⁸ = 0
the condition that must be followed is
(x − 4)³ = 0 ..........(1)
or
(x − 2)⁸ = 0 ..........(2)
considering the equation 1, we have
(x − 4)³ = 0
or
x -4 = 0
or
x = 4
and,
for (x − 2)⁸ = 0
or
x - 2 = 0
or
x = 2
Thus,
zeros of the function are x = 2 and, x = 4
c) h(x) = (2x − 3)⁵
now,
for the (2x − 3)⁵ = 0
the condition that must be followed is
(2x − 3)⁵ = 0
or
2x - 3 = 0
or
2x = 3
or
x =
Thus,
zeros of the function are x =
d) k(x) =(3x +4)¹⁰⁰(x − 17)⁴
now,
for the (3x +4)¹⁰⁰(x − 17)⁴ = 0
the condition that must be followed is
(3x +4)¹⁰⁰ = 0 ..........(1)
or
(x − 17)⁴ = 0 ..........(2)
considering the equation 1, we have
(3x +4)¹⁰⁰ = 0
or
(3x +4) = 0
or
3x = -4
or
x =
and,
for (x − 17)⁴ = 0
or
x - 17 = 0
or
x = 17
Thus,
zeros of the function are x = and, x = 17