total number of roots of (x^2+x+1)^2 +2 = (x^2+x+1)(x^2-2x-6)
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Answer: The original equation has NO real roots.
Explanation : x
2
+
1
)
2
−
x
2
=
0
(
x
2
+
1
−
x
)
(
x
2
+
1
+
x
)
=
0
x
2
−
x
+
1
=
0
∨
x
2
+
x
+
1
=
0
x
2
−
x
+
1
=
0
Δ
=
(
−
1
)
2
−
4
⋅
1
⋅
1
=
1
−
4
=
−
3
This equation has NO real roots.
x
2
+
x
+
1
=
0
Δ
=
1
2
−
4
⋅
1
⋅
1
=
1
−
4
=
−
3
This equation has NO real roots.
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