The number of real or complex solutions of `x^(2)-6|x|+8=0` is

Options

A. `6`
B. `7`
C. `8`
D. `9`

Answer

A

Explanation

`(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.