This downside comes from Oxford’s 2023 MAT.
What number of distinct actual options are there to the equation:
(((x2 – 1)2 – 2)2 – 3)2 = 4
As typical, watch the video for an answer.
Oxford Entrance Examination How Many Distinct Actual Options
Or hold studying.
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Reply To Oxford Entrance Examination How Many Distinct Actual Options
(Just about all posts are transcribed shortly after I make the movies for them–please let me know if there are any typos/errors and I’ll right them, thanks).
Begin from the surface. The sq. of an actual quantity is non-negative, so any time a amount squares to be a damaging worth that risk can not yield an actual answer.
(((x2 – 1)2 – 2)2 – 3)2 = 4
((x2 – 1)2 – 2)2 – 3 = -2 or 2
((x2 – 1)2 – 2)2 = 1 or 5
(x2 – 1)2 – 2 = -1 or 1 or -√5 or √5
(x2 – 1)2 = 1 or 3 or 2 – √5 or 2 + √5
Since 2 – √5 < 0, it won’t correspond to an actual valued answer for x.
(x2 – 1)2 = 1 or 3 or 2 + √5
x2 – 1 = -1 or 1 or -√3 or √3 or √(2 + √5) or -√(2 + √5)
x2 = 0 or 2 or 1 – √3 or 1 + √3 or 1 + √(2 + √5) or 1 – √(2 + √5)
Each 1 – √3 and 1 – √(2 + √5) are damaging so get rid of them.
x2 = 0 or 2 or 1 + √3 or 1 + √(2 + √5)
x = 0 or ±√2 or ±√(1 + √3) or ±√(1 + √(2 + √5))
Thus there are 7 distinct actual roots.
References
Oxford MAT
https://www.maths.ox.ac.uk/system/recordsdata/attachments/websolutions23_1.pdf
https://www.maths.ox.ac.uk/system/recordsdata/attachments/test23.pdf
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