1 3 4 9 x 2 5 4 = = = = = = a + b + c + d + e 2 a + 4 b + 8 c + 1 6 d + 3 2 e 3 a + 9 b + 2 7 c + 8 1 d + 2 4 3 e 4 a + 1 6 b + 6 4 c + 2 5 6 d + 1 0 2 4 e 5 a + 2 5 b + 1 2 5 c + 6 2 5 d + 3 1 2 5 e 6 a + 3 6 b + 2 1 6 c + 1 2 9 6 d + 7 7 7 6 e
What is x ?
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You can have 5 rows of differences if you include the evaluation of the function at x = 0. This would leave your table unchanged but would add the following leading entry in each of the five rows of differences in your table. Row 1: 0, Row 2: 1, Row 3: 1, Row 4: -2, Row 5: 7. Now we can add an additional row, having two entries: (x - 30) , and (290 - 5 x). Now these two are equal. So, 6 x = 3 2 0 , Therefore, x = 5 3 3 1 ,
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Thank you for pointing out. I would've never noticed if you were not to tell me.
I edited the problem. Now it should be all good.
Again, I appreciate your report!
Brilliant staff, could you give clarification to Hajjir statement?
But you did not repeat the difference calculations 5 times. You did only 4 rows of differences.
The actual answer is 53.33333, i.e. 5 3 3 1
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Seems like you'll need a lot of calculations, but actually you calculate only 2 1 times.
Let's define f ( x ) = a x + b x 2 + c x 3 + d x 4 + e x 5 .
Then f ( 0 ) = 0 , f ( 1 ) = 1 , f ( 2 ) = 3 , f ( 3 ) = 4 , f ( 4 ) = 9 , f ( 5 ) = x , f ( 6 ) = 2 2 6 .
Generally, performing the calculation f ( n + 1 ) − f ( n ) decreases the degree of the function by 1 .
Since f ( x ) is a function of 5th degree, it becomes an invariable when we perform the calculation 5 times.
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First, I'll let you know how I expressed the calculation.
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a b − a b c − 2 b + a c − b c
Good, now let's begin.
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0 1 1 1 2 − 2 3 − 1 7 1 5 x − 3 0 4 4 x − 2 3 5 x − 1 8 3 1 8 − 5 x 9 x − 1 4 2 9 5 − 4 x x − 9 2 7 7 − 3 x x 2 6 3 − 2 x 2 5 4 − x 2 5 4
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x − 3 0 = 3 1 8 − 5 x ,
∴ x = 5 8