Question
How many distinct real roots are possible for the below equation?
x6−26x3−27=0
x6−26x3−27=0
Answer: Option B
:
B
Let x3=y
y2−26y−27=0
y2−27y+y−27=0
y(y−27)+1(y−27)=0
(y+1)(y−27)=0
The roots are y = -1, y = 27
x3=−1,x=−1
x3=27,x=3
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:
B
Let x3=y
y2−26y−27=0
y2−27y+y−27=0
y(y−27)+1(y−27)=0
(y+1)(y−27)=0
The roots are y = -1, y = 27
x3=−1,x=−1
x3=27,x=3
Was this answer helpful ?
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