x 3 − 4 5 3 x 2 + 6 8 4 0 2 x − 3 4 4 2 8 0 0
If the expression above has 3 real roots r 1 , r 2 , r 3 in an arithmetic progression having common difference 1 then find the value of
r 1 2 + r 2 3 + r 3 2 .
Note : The exponents in the expression are indeed 2 , 3 , 2 .
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A more simplified solution Consider the roots be
a − 1 , a , a + 1
a − 1 + a + a + 1 = 4 5 3
From here
a = 1 5 1
Hence
1 5 0 2 + 1 5 1 3 + 1 5 2 2 = 3 4 8 8 5 5 5
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Good solution.
same as mine ! :)
Exactly same solution
I did the same way.
The roots of your polynomial are NOT 150, 151, 152.
Assuming that the polynomial is correct, then r 1 2 + r 2 2 + r 3 2 = ( r 1 + r 2 + r 3 ) 2 − 2 ( r 1 r 2 + r 2 r 3 + r 3 r 1 ) = 4 5 3 2 − 2 × 6 8 4 0 2 = 6 8 4 0 5 .
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I have updated question. Now roots are 150,151,152
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I apologize for the error.
I did not notice that the term r 2 in the expression has a cube in it. I have updated the question to reflect that. 3488555 is the correct answer.
Sorry for the confusion that this caused.
Calvin, the roots are 150,151 & 152. Why do you say they are not so?
since the sum of the roots = 453 and the first root = 150, then to become in AP, The roots are 150,151 and 152
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Sum of roots by vieta's formula is 4 5 3 .
We know AP sum formula S n = 2 n ( 2 a + ( n − 1 ) d ) )
In this formula we have n = 3 , a = r 1 , S 3 = r 1 + r 2 + r 3
Therefore
S 3 = 2 3 ( 2 a + ( 2 ) × 1 )
S 3 = 3 a + 3
We know sum of roots is 4 5 3
Therefore
4 5 3 = 3 a + 3
⇒ a = 1 5 0 = r 1
⇒ r 2 = 1 5 1
⇒ r 3 = 1 5 2
⇒ r 1 2 + r 2 3 + r 3 2 = 1 5 0 2 + 1 5 1 3 + 1 5 2 2 = 2 2 5 0 0 + 3 4 4 2 9 5 1 + 2 3 1 0 4 = 3 4 8 8 5 5 5