InterviewSolution
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InterviewSolution
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The number of real or complex solutions of `x^(2)-6|x|+8=0` isA. `6`B. `7`C. `8`D. `9` |
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Answer» Correct Answer - A `(a)` If `x` is real, `x^(2)-6|x|+8=0` `implies |x|^(2)-6|x|+8=0` `implies |x|=2,4` `implies x=+-2,+-4` If `x` is non-real, say `x=alpha+I beta`, then `(alpha+I beta)^(2)-6sqrt(alpha^(2)+beta^(2))+8=0` `:.alpha^(2)-beta^(2)+8-6sqrt(alpha^(2)+beta^(2))+2i alpha beta=0` Comparing real and imaginary parts, `alpha beta=0 implies alpha=0` (if `beta=0`, then `x` is real) and `-beta^(2)+8-66sqrt(beta^(2))=0` `:. beta+-6beta-8=0` `implies beta=(overset(-)(+)6overset(-)(+)sqrt(68))/(2)` `implies beta=+-(3-sqrt(17))` Hence, `+-(3-sqrt(17))i` are non-real roots. |
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