Both equations x^2 + b^2 = 1 - 2bx and x^2 + | vedclass

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(D) The given equations can be rewritten as:$x^2 + 2bx + b^2 = 1 \implies (x + b)^2 = 1 \implies x = -b \pm 1$$x^2 + 2ax + a^2 = 1 \implies (x + a)^2

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(D) The given equations can be rewritten as:$x^2 + 2bx + b^2 = 1 \implies (x + b)^2 = 1 \implies x = -b \pm 1$$x^2 + 2ax + a^2 = 1 \implies (x + a)^2 = 1 \implies x = -a \pm 1$Since each equation has exactly one root,this implies the discriminant is zero,but here the equations are already perfect squares. The condition that they share a root means one of the roots from the first set must equal one of the roots from the second set.Given $a 
eq b$,the roots are $\{-b+1, -b-1\}$ and $\{-a+1, -a-1\}$.For them to share a root,we must have:$-b + 1 = -a - 1 \implies a - b = -2$$OR$$-b - 1 = -a + 1 \implies a - b = 2$Both cases are covered by the condition $|a - b| = 2$.Since $|a - b| = 2$ implies $a - b = 2$ or $a - b = -2$,and $a - b + 2 = 0$ is equivalent to $a - b = -2$,all options $A, B,$ and $C$ are correct.

Both equations $x^2 + b^2 = 1 - 2bx$ and $x^2 + a^2 = 1 - 2ax$ have exactly one root each,and they share the same root. Then:

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