Hop, skip and a jump

Algebra Level 3

Let a , b , c a,b,c be the roots of the equation x 3 + 3 x 2 + 7 x + 6 = 0. x^{3} + 3x^{2} + 7x + 6 = 0.

Find ( a b ) 2 + ( b c ) 2 + ( a c ) 2 . (ab)^{2} + (bc)^{2} + (ac)^{2}.


The answer is 13.

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Brian Charlesworth by Brian Charlesworth , 55, Canada

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3 solutions

By Vieta's rule we have that a + b + c = − 3 , a b + b c + a c = 7 a + b + c = -3, ab + bc + ac = 7 and a b c = − 6. abc = -6.

So ( a b ) 2 + ( b c ) 2 + ( a c ) 2 = ( a b + b c + a c ) 2 − 2 a b c ( a + b + c ) = (ab)^{2} + (bc)^{2} + (ac)^{2} = (ab + bc + ac)^{2} - 2abc(a + b + c) =

7 2 − 2 ( − 6 ) ( − 3 ) = 13 . 7^{2} - 2(-6)(-3) = \boxed{13}.

16 Helpful 0 Interesting 1 Brilliant 1 Confused

I did the same . Upvoted :)

Nihar Mahajan - 6 years, 3 months ago

what is the meaning of the title of the problem??

Kislay Raj - 6 years, 3 months ago

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With titles I sometimes just write the first thing that pops into my head. I guess I was thinking of the three terms found using Vieta's rule and how they were used in solving for the given expression; just a "hop, skip and a jump" to a solution. The phrase is normally used to indicate a short distance, or a relatively simple task. The problem, however, could be quite difficult if one doesn't know Vieta's rule and see the factoring 'trick' right away.

Brian Charlesworth - 6 years, 3 months ago

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And that's why the problem is level 4!! , where actually its a very simple and easy one. :)

Kislay Raj - 6 years, 3 months ago

I did exactly the same!!! :)

VIDIT LOHIA - 6 years, 3 months ago

I used binomial theorem to expand assuming that (ab+bc) is one term then expanding them again Is there a rule for the expansion you did ? @brian charlesworth

Muhammad Ahmad - 6 years, 3 months ago

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I don't think that there is any rule for this expansion; I guess that it's more about just looking for a way of using the Vieta's "components" to create the given expression.

Brian Charlesworth - 6 years, 3 months ago
Guru Prasaadh
Mar 4, 2015

this ones an overrated problem

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Using Vieta's Rule, we have that a + b + c = − 3 a+b+c=-3 , a b + b c + c a = 7 ab+bc+ca=7 and a b c = − 6 abc=-6 .

Hence, ( a b ) 2 + ( b c ) 2 + ( c a ) 2 + 2 a b c ( a + b + c ) = ( a b + b c + c a ) 2 ⇔ ( a b ) 2 + ( b c ) 2 + ( c a ) 2 = ( a b + b c + c a ) 2 − 2 a b c ( a + b + c ) ⇔ ( a b ) 2 + ( b c ) 2 + ( c a ) 2 = 7 2 − 2 × ( − 6 ) × ( − 3 ) = 13 { (ab) }^{ 2 }+{ (bc) }^{ 2 }+{ (ca) }^{ 2 }+2abc(a+b+c)={ (ab+bc+ca) }^{ 2 }\\ \Leftrightarrow { (ab) }^{ 2 }+{ (bc) }^{ 2 }+{ (ca) }^{ 2 }={ (ab+bc+ca) }^{ 2 }-2abc(a+b+c)\\ \Leftrightarrow { (ab) }^{ 2 }+{ (bc) }^{ 2 }+{ (ca) }^{ 2 }={ 7 }^{ 2 }-2\times (-6)\times (-3)= \boxed{13}

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a b c ≠ 6 abc \neq 6 , a b c = − 6 abc = -6

Nihar Mahajan - 6 years, 3 months ago

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Thanks for the correction. I've been too lazy to check it, sorry :P

Trung Đặng Đoàn Đức - 6 years, 3 months ago

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