The roots of the equation (a+b)x2−15+(a−b)x2−15=2a , where a2−b=1, are

We have,a−b=(a−b)(a+b)a+b=a2−ba+b=1a+b[∵a2−b=1]So, by putting (a+b)x2−15=y, the given equation becomes  y+1y=2a⇒y2−2ay+1=0⇒(y−a)2=a2−1⇒y−a=±a2−1⇒y−a=±b⇒y=a±b⇒(a+b)x2−15=a+b,a−b⇒x2−15=1or,x2−15=−1⇒x=±4,x=±14.