The roots of the equation (a+b)x2−15+(a−b)x2−15=2a, where a2−b=1, are
±2,±3
±4,±14
±3,±5
±6,±20
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 ∵a2−1=b⇒ y=a±b⇒ (a+b)x2−15=a+b,a−b
⇒ x2−15=1 or, x2−15=−1⇒x=±4,x=±14
ALITER We have,
(a+b)x2−15+(a−b)x2+15=(a+a)1+(a−b)1⇒ x2−15=±1⇒x=±4,x=±14