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Determine the value of real parameter $p$

in such a way that the equation

$$\sqrt{x^2+2p} = p+x $$

has just one real solution

a. $p \ne 0$

b. There is no such value of parameter$p$

c. None of the remaining possibilities is correct.

d. $p\in [−2,\infty)$

e. $p\in [−2,0)\cup(0,\infty)$

I thought the answer is A but it isn't. help me!

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The condition says $x+p\geq0$.

$x^2+2p=(x+p)^2$

$⇒x=1-\dfrac p2$

from condition

$1+\dfrac p2\geq0$

$⇒p\geq-2$

Then consider the case of $p=0$, this has all x are solution.

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We have $$\begin{align}\sqrt{x^2+2p} = p+x&\implies x^2+2p=p^2+2px+x^2\\&\implies 2p-p^2=2px\end{align}$$ and if $p\neq0$ then $$x=\frac{2-p}{2}$$ which is the only solution.

If $p=0$ then we get $0=0$ which implies that there are infinitely many solutions.

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