find all the zeros of x^2-18
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Answer:
Complete the square then use the difference of squares identity:
a
2
−
b
2
=
(
a
−
b
)
(
a
+
b
)
with
a
=
(
x
+
3
)
and
b
=
3
i
as follows:
f
(
x
)
=
x
2
+
6
x
+
18
=
x
2
+
6
x
+
9
+
9
=
(
x
+
3
)
2
+
3
2
=
(
x
+
3
)
2
−
(
3
i
)
2
=
(
(
x
+
3
)
−
3
i
)
(
(
x
+
3
)
+
3
i
)
=
(
x
+
3
−
3
i
)
(
x
+
3
+
3
i
)
Hence
f
(
x
)
=
0
when
x
=
−
3
±
3
i
Answer:
x^2 - 18 = 0
=> x^2 = 18
=> x = root 18
=> x = 4.24
Zero of the polynomial is 4.24
Hope it helps.