Can Vieta help?

Algebra Level 4

x 3 453 x 2 + 68402 x 3442800 \large { x }^{ 3 }-{ 453x }^{ 2 }+68402x-3442800

If the expression above has 3 real roots r 1 , r 2 , r 3 { 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 . { { r }_{ 1 } }^{ 2 }+{ { r }_{ 2 } }^{ 3 }+{ { r }_{ 3 } }^{ 2 }.

Note : The exponents in the expression are indeed 2 , 3 , 2 2, 3, 2 .


The answer is 3488555.

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2 solutions

Shivamani Patil
Sep 29, 2014

Sum of roots by vieta's formula is 453 453 .

We know AP sum formula S n = n 2 ( 2 a + ( n 1 ) d ) { S }_{ n }=\frac { n }{ 2 } \left( 2a+\left( n-1 \right) d \right) )

In this formula we have n = 3 n=3 , a = r 1 , S 3 = r 1 + r 2 + r 3 a={ r }_{ 1 } ,{ S }_{ 3 }={ r }_{ 1 }+{ r }_{ 2 }+{ r }_{ 3 }

Therefore

S 3 = 3 2 ( 2 a + ( 2 ) × 1 ) { S }_{ 3 }=\frac { 3 }{ 2 } \left( 2a+\left( 2 \right) \times 1 \right)

S 3 = 3 a + 3 { S }_{ 3 }=3a+3

We know sum of roots is 453 453

Therefore

453 = 3 a + 3 453=3a+3\quad

a = 150 = r 1 \Rightarrow a=150={ r }_{ 1 }

r 2 = 151 \Rightarrow { r }_{ 2 }=151

r 3 = 152 \Rightarrow { r }_{ 3 }=152

r 1 2 + r 2 3 + r 3 2 = 150 2 + 151 3 + 152 2 = 22500 + 3442951 + 23104 = 3488555 \Rightarrow { { r }_{ 1 } }^{ 2 }+{ { r }_{ 2 } }^{ 3 }+{ { r }_{ 3 } }^{ 2 }={ 150 }^{ 2 }+{ 151 }^{ 3 }+{ 152 }^{ 2 }=22500+3442951+23104=3488555

A more simplified solution Consider the roots be

a 1 a-1 , a a , a + 1 a+1

a 1 + a + a + 1 = 453 a-1 + a + a+1 = 453

From here

a = 151 a = 151

Hence

15 0 2 + 15 1 3 + 15 2 2 = 3488555 150^{2} + 151^{3} + 152^{2} = \boxed{3488555}

Krishna Sharma - 6 years, 8 months ago

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Good solution.

shivamani patil - 6 years, 8 months ago

same as mine ! :)

Nihar Mahajan - 6 years, 4 months ago

Exactly same solution

Kushagra Sahni - 5 years, 11 months ago

I did the same way.

Niranjan Khanderia - 5 years, 7 months ago

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 ) = 45 3 2 2 × 68402 = 68405. 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 ) = 453^2 - 2 \times 68402 = 68405.

Calvin Lin Staff - 6 years, 8 months ago

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I have updated question. Now roots are 150,151,152

shivamani patil - 6 years, 8 months ago

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I apologize for the error.

I did not notice that the term r 2 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 Lin Staff - 6 years, 8 months ago

Calvin, the roots are 150,151 & 152. Why do you say they are not so?

ajit athle - 1 year, 10 months ago
Fox To-ong
Feb 13, 2015

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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