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Bunuel
If \(\sqrt{3-x}= \sqrt{x} + 3\), then \(x^2=\)

(A) 1
(B) 3
(C) 2 − 3x
(D) 3x − 1
(E) 3x − 9

Let's square each side of the given equation:

\(\Rightarrow\) (√(3 - x))^2 = (√x + 3)^2

\(\Rightarrow\) 3 - x = x + 6√x + 9

\(\Rightarrow\) -6 - 2x = 6√x

\(\Rightarrow\) -3 - x = 3√x

Let's square each side of the equation again:

\(\Rightarrow\) (-3 - x)^2 = (3√x)^2

\(\Rightarrow\) 9 + 6x + x^2 = 9x

\(\Rightarrow\) x^2 = 3x - 9

Answer: E
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↧↧↧ Detailed Video Solution to the Problem Series ↧↧↧



\(\sqrt{3-x}= \sqrt{x} + 3\)

Squaring both the sides we get

\((\sqrt{3-x})^2= (\sqrt{x} + 3)^2\)
=> \(3 - x = (\sqrt{x})^2 + 2 * \sqrt{x} * 3 + 3^2\)
=> 3 - x = x + 6\(\sqrt{x}\) + 9
=> 3 - x - x - 9 = 6\(\sqrt{x}\)
=> -2x - 6 = 6\(\sqrt{x}\)
=> -(x + 3) = 3\(\sqrt{x}\)

Squaring both the sides we get
=> \((x + 3)^2 = (3\sqrt{x})^2\)
=> \(x^2 + 2 * x * 3 + 3^2 = 9x\)
=> \(x^2\) + 6x + 9 = 9x
=> \(x^2\) = 9x - 6x - 9 = 3x - 9

So, Answer will be E
Hope it helps!

Watch the following video to MASTER Roots

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