If $x^{2} - 1 = - b^{2} - 2 b x a n d x^{2} - 1 = - a^{2} - 2 a x$ have exactly one root in common then

a = b + 2

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$a = b^{2} + 2$

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a + b+ 2 = 0

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a - b + 2 = 0

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Solution

The correct option is A
a = b + 2

$\begin{matrix} x^{2} + b^{2} + 2 b x = 1 & x^{2} + a^{2} + 2 a x = 1 \Rightarrow (x + b)^{2} = 1 & (x + a)^{2} = 1 \Rightarrow x + b = \pm 1 & x + a = \pm 1 \Rightarrow x = - b + 1, - b - 1 & x = - a + 1, - a - 1 \end{matrix}$
-b+1=-a+1, -b+1=-a-1 -b-1=-a+1, -b - 1 = -a -1
a = b, a = 2 - b, a - b = 2, a = b.
$\Rightarrow$ a = 2 + b
Now a = b is incorrect ( $∵$ both roots are identical)

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Similar questions

Given two quadratic equations $x^{2} - 1 = - b^{2} - 2 b x & x^{2} - 1 = - a^{2} - 2 a x$ have exactly one root in common then select the correct statements.

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If x2−1=−b2−2bx and x2−1=−a2−2ax have exactly one root in common then

The correct option is A a = b + 2 x2+b2+2bx=1x2+a2+2ax=1⇒(x+b)2=1(x+a)2=1⇒x+b=±1x+a=±1⇒x=−b+1, −b−1x=−a+1, −a−1 -b+1=-a+1, -b+1=-a-1 -b-1=-a+1, -b - 1 = -a -1 a = b, a = 2 - b, a - b = 2, a = b. ⇒ a = 2 + b Now a = b is incorrect (∵ both roots are identical)

If $x^{2} - 1 = - b^{2} - 2 b x a n d x^{2} - 1 = - a^{2} - 2 a x$ have exactly one root in common then