If \( \Large \sqrt{3x^{2}-7x-30} + \sqrt{2x^{2}-7x-5} = x+5 \), then x is equal to:
Correct Answer: Description for Correct answer:
We have
\( \Large \sqrt{3x^{2}-7x-30}+\sqrt{2x^{2}-7x-5}=x+5 \)
or \( \Large \sqrt{3x^{2}-7x-30}= \left(x+5\right)-\sqrt{2x^{2}-7x-5} \)
on squaring both sides, we get
\( \Large 3x^{2}-7x-30=x^{2}+25+10x+ \left(2x^{2}-7x-5\right) \)
\( \Large -2 \left(x+5\right)\sqrt{2x^{2}-7x-5} => \sqrt{2x^{2}-7x-5}=5 \)
Again squaring
=> \( \Large 2x^{2}-7x-30=0 => x=6 \)
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