solve 12x^4-56x^3+89x^2-56x+12=0
Answers
Answered by
2
Answer:
mark brainlist plzplzplzplz
Step-by-step explanation:
Given
12x
4
−56x
3
+89x
2
−56x+12=0 ----- (1)
on dividing equation (1) by x
2
⇒12x
2
−56x+89−
x
56
+
x
2
12
=0
12[x
2
+
x
2
1
]−56[x+
x
1
]+89=0
⇒12[(x+
x
1
)
2
−2]−56[x+
x
1
]+89=0
⇒12[x+
x
1
]
2
−56[x+
x
1
]+89−24=0
⇒12[x+
x
1
]
2
−56[x+
x
1
]+65=0
Let x+
x
1
=y
12y
2
−56y+65=0
y=
24
56+
(56)
2
−48×65
=
6
13
,
2
5
Now x+
x
1
=
6
13
or x+
x
1
=
2
5
6x
2
−13x+6=0 or 2x
2
−5x+2=0
x=
12
13±
169−144
or x=
4
5±
25−16
x=
12
13+5
or x=
4
5+3
Hence [x=
2
3
], or [
3
2
] [x=2,
2
1
]
Answered by
5
Answer:
14 this is right answer
Similar questions