Question

Please solve this question

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Answer 1

Answer:

$\sf x = a\ or\ x = \frac{a}{2}$

Step-by-step explanation:

$\sf \sqrt{a^2 + 2ax - 3x^2} - \sqrt{a^2 + ax - 6x^2} = \sqrt{2a^2 + 3ax - 9x^2}$

It can be written as :-

$\sf \sqrt{-3x^2 + 2ax+a^2} - \sqrt{-6x^2 + ax +a^2} = \sqrt{-9x^2 + 3ax +2a^2}$

Now split the middle terms :-

$\sf \sqrt{-3x^2 + 2ax+a^2} - \sqrt{-6x^2 + ax +a^2} = \sqrt{-9x^2 + 3ax +2a^2}\\\\\tiny{\sf \sqrt{-3x^2 + 3ax-ax+a^2} - \sqrt{-6x^2 + 3ax - 2ax+a^2} = \sqrt{-9x^2 + 6ax-3ax +2a^2}}$

Now, after taking out the common factors, expression would be like :-

$\sf\sqrt{(-3x-a)(x-a)}-\sqrt{(-3x-a)(2x-a)}=\sqrt{(-3x-a)(3x-2a)}$

taking √(-3x -a) as common, we get

$\sf\sqrt{(-3x-a)}(\sqrt{x-a}-\sqrt{2x-a}) = (\sqrt{-3x-a})(\sqrt{3x-2a)}$

cancel √(-3x - a) from both sides

$\sf \implies \sqrt{x-a} - \sqrt{2x-a} = \sqrt{3x-2a}$

Now, square both the sides

$\sf x -a + 2x -a - 2\sqrt{(x -a)(2x-a)} = 3x - 2a \\\\Using \ the \ identity \ (a - b)^2 = a^2+b^2-2ab)$

Transposing the root on other side and like terms on the other :-

$\sf x -a + 2x -a - 2\sqrt{(x -a)(2x-a)} = 3x - 2a\\\\\implies x - a +2x - a- 3x + 2a= 2\sqrt{(x-a)(2x-a)}\\\\\implies 3x -3x -2a + 2a = 2\sqrt{(x-a)(2x-a)}\\\\\implies 0 = 2\sqrt{(x-a)(2x-a)}\\\\\implies \sqrt{(x-a)(2x-a)}=\frac{0}{2}\\\\\implies \sqrt{(x-a)(2x-a)}= 0$

again, square both sides

$\sf\implies (x -a)(2x-a) = 0 \\\\So, either\ x - a = 0 \ or \ 2x - a = 0 \\\\\implies x = a\ or\ 2x = a \\\\\implies x = a\ or\ x = \frac{a}{2}$

Hence, x = a or x = a/2