The number of solutions of (5+2√6)x2−3+(5−2√6)x2−3=10 are
A
0
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B
2
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C
4
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D
6
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Solution
The correct option is C4 (5+2√6)x2−3+(5−2√6)x2−3=10 Let t=(5+2√6)x2−3, then (5−2√6)x2−3=1t ⇒t+1t=10 ⇒t2−10t+1=0 ⇒t=10+√962 or t=10−√962 ⇒t=5+2√6 or t=5−2√6 ⇒x2−3=1 or x2−3=−1 Hence, x can take four real values.