Solution 2.2:3d

From Förberedande kurs i matematik 1

(The exercise is taken from an actual exam in Spring Term 1945!)

There are no common factors on the left-hand side which we can take out, so we choose to expand the three terms on the left-hand side:

x2−314x+21=x214x−x221−314x−321=12x2+x1−34x−23=12x2+14x−2312x−322=12x2−212x32+322=14x2−23x+94

12x+3112x−31=conjugate rule=12x2−132=14x2−91 

Collecting up terms, the left-hand side becomes

12x2+14x−23−14x2−23x+94−14x2−91=21−41−411x2+41+32x1+−23−94+91=42−1−11x2+343+24x1+29−39−42+12=1134x1−3329

and because 33=311, 9=33 and 4=22, the whole equation can rewritten as

11322x1−311233=0

Taking out common factors, we get

113212x−1=0 

and then we see that the equation has the solution x=12.

Finally, we substitute x=12 into the original equation to check that we have calculated correctly.

221−31421+21−1221−322−1221+311221−31=4−321+21−1−322−1+311−31=1−312−3432=1−91−98=0