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I am stuck on this problem:

  1. If $\sqrt{x^2+4ax+5}+\sqrt{x^2+4bx+5}=2(a-b)$ then $x=?$

I did the following things

1) For solution to exist a≥b

2) $x^2+4ax+5≥0$ which implies that $16a^2-20≤0$

On differentiating the quadratic $x^2+4ax+5=0$, I get $x=-2a$. Since the parabola is opening upwards the function must lie above or on the x axis at thus x. Therefore, $4a^2-8a+5≥0$

From both of these I got the range of a,b as $R-[-1/2,1/2]$

Beyond this, I am stuck. Any hints?

1 Answers 1

1

We have, on simple manipulation, $$\sqrt {x^2+4ax+5} + \sqrt {x^2+4bx+5} =2 (a-b) $$ $$\Rightarrow x^2+4ax+5 =4 (a-b)^2+ x^2+4bx +5 -4 (a-b)\sqrt {x^2+4bx+5} $$ $$\Rightarrow 4ax -4bx =4 (a-b)(a-b) -4 (a-b)\sqrt {x^2+4bx+5} $$

Assuming $(a-b) \neq 0$, we get, $$x = (a-b) -\sqrt {x^2+4bx+5} $$ $$ \Rightarrow x^2+4bx+5 = (a-b)^2 +x^2 -2x (a-b) $$ $$ \Rightarrow 2ax + 2bx = (a-b)^2+5$$ $$\Rightarrow x = ? $$

Hope it helps.