
To solve : the equation

Answer to Problem 21PPS
The equation has no solution
Explanation of Solution
Given information :
The equation is 2y−5+y−12y+1=22y2−9y−5
Formula used : Sridhar Acharya
−b±√b2−4ac2a
Calculation :
The LCD for the term is (y−5)(2y+1) .
2y−5+y−12y+1=22y2−9y−5(y−5)(2y+1)(2y−5)+(y−5)(2y+1)(y−12y+1)=(y−5)(2y+1)(22y2−9y−5)2(2y+1)+(y−1)(y−5)=(2y2+10x+y−5)(22y2−9y−5)4y+2+y2−y−5y+5=(2y2−9y−5)(22y2−9y−5)y2−2y+7=2y2−2y+7−2=0y2−2y+5=0y=−b±√b2−4ac2ay=−(−2)±√(−2)2−4⋅1⋅52⋅1y=2±√4−202y=2±√−162y=2±4√12y=1±2√1y=1±2i
Check 2y−5+y−12y+1=22y2−9y−5
Take positive sing
21+2i−5+1+2i−12(1+2i)+i=22(1+2i)2−9(1+2i)−522i−4+2i2+4i+1=22{(1)2+2(1)(2i)+(2i)2}−9−18i−522(i−2)+2i3+4i=22(1+4i+4)−14−18i1i−2+2i3+4i=15+4i−9i−73+4i+2i(i−2)(i−2)(3+4i)=1−5i−23+4i+2i(i)−4i3i−6+4i(i)−8i=1−5i−23+2i2−5i−6+4i2=1−5i−23+2−5i−6+4=1−5i−25−5i−2=1−5i−2
Take negative sing.
21−2i−5+1−2i−12(1−2i)+i=22(1−2i)2−9(1−2i)−52−2i−4−2i2−4i+1=22{(1)2−2(1)(2i)+(2i)2}−9+18i−522(−i−2)−2i3−4i=1(1−4i+4i2)+9i−71−i−2−2i3−4i=1(1−4i+4i)+9i−7(3−4i)−2i(−i−2)(3−4i)(−i−2)=15−4i+9i−73−4i+2i2+4i−3i+4i−6+8i=15i−23+2−5i−6+4=15i−25−5i−2=15i−2
Hence, the equation has no solution.
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