The value of x  satisfying the equation (5+26)x2−3+(5−26)x2−3=10  are

The given equation is (5+26)x2−3+(5−26)x2−3=10 5−26=(5−26)(5+26)5+26       ( rationalization)                =25−245+26=15+26 Let (5+26)x2−3=y The given equation becomesy+1y=10   ⇒y2−10y+1=0 y=10±100−42=10±962=10±462=5±26 When y=5+26 (5+26)x2−3=5+26 ∴ x2−3=1  ⇒x2=4  ⇒x=±2 When y=5−26 (5+26)x2−3=5−26=(5+26)−1 ∴x2−3=−1  ⇒x2=2 ⇒x=±2