Solution 2.2:2d
From Förberedande kurs i matematik 1
First, we move all the terms over to the left-hand side
x2+4x+1
2+3x4−2x2−
2x2+2x+3
2=0
As the equation stands now, it seems that the best approach for solving the equation is to expand the
squares, simplify and see what it leads to.
When the squares are expanded, each term inside a square is multiplied by itself and all other terms:
x2+4x+1
2=
x2+4x+1
x2+4x+1
=x2
x2+x2
4x+x2
1+4x
x2+4x
4x+4x
1+1
x2+1
4x+1
1=x4+4x3+x2+4x3+16x2+4x+x2+4x+1=x4+8x3+18x2+8x+1
2x2+2x+3
2=
2x2+2x+3
2x2+2x+3
=2x2
2x2+2x2
2x+2x2
3+2x
2x2+2x
2x+2x
3+3
2x2+3
2x+3
3=4x4+4x3+6x2+4x3+4x2+6x+6x2+6x+9=4x4+8x3+16x2+12x+9
After we collect together all terms of the same order, the left hand side becomes
x2+4x+1
2+3x4−2x2−
2x2+2x+3
2=
x4+8x3+18x2+8x+1
+3x4−2x2−
4x4+8x3+16x2+12x+9
=
x4+3x4−4x4
+
8x3−8x3
+
18x2−2x2−16x2
+
8x−12x
+
1−9
=−4x−8
After all simplifications, the equation becomes
x=−2
Finally, we check that
−2
2+4
−2
+1
2+3
−2
4−2
−2
2=
4−8+1
2+3
16−2
4=
−3
2+48−8=9+48−8=49
2
−2
2+2
−2
+3
2=
2
4−4+3
2=72=49