Given two quadratic equations x2−1=−b2−2bx & x2−1=−a2−2ax have exactly one root in common then select the correct statements.
a = b + 2
Given Quadratic Equations:
x2−1=−b2−2bx⇒x2+b2+2bx=1⇒(x+b)2=1⇒(x+b)=±1⇒x=1−b or x=−1−b
And similarly for other equation:
x2−1=−a2−2ax⇒x2+a2+2ax=1⇒(x+a)2=1⇒(x+a)=±1⇒x=1−a or x=−1−a
Both the equation have one common root:
∴ The possible cases are:
A.1−b=1−a⇒a=b
B.1−b=−1−a⇒a−b=−2
C.−1−b=1−a⇒a−b=2
D.−1−b=−1−a⇒a=b
Now, the condition a=b is to be rejected as it will mean that both the roots are equal.
Thus a−b=−2 & a−b=2