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Algebra Level 4

Let w w denote a non-real complex number where w 7 = 1 w^7 = 1 . Define a = w + w 2 + w 4 a = w+w^2+w^4 and b = w 3 + w 5 + w 6 b = w^3 + w^5 + w^6 .

If a a and b b are roots of the equation x 2 + p x + q = 0 x^2+px+q=0 for constants p p and q q , what is the value of 10 p + q 10p + q ?


The answer is 12.

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De Azalea by de azalea , 24, Indonesia

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2 solutions

By Vieta's formulas, we know that p = − ( a + b ) p=-(a+b) and q = a b q=ab , so:

p = − ( w + w 2 + w 4 + w 3 + w 5 + w 6 ) = 1 q = ( w + w 2 + w 4 ) ( w 3 + w 5 + w 6 ) = w 4 + w 6 + 1 + w 5 + 1 + w + 1 + w 2 + w 3 = 2 p=-(w+w^2+w^4+w^3+w^5+w^6)=1 \\ q=(w+w^2+w^4)(w^3+w^5+w^6)=w^4+w^6+1+w^5+1+w+1+w^2+w^3=2

So, 10 p + q = 10 ( 1 ) + 2 = 12 10p+q=10(1)+2=\boxed{12}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Since w is a root in the 7th cyclotomic polynomial we see that a+b=-1 and ab=2. By viete's formulae we get that the polynomial is x 2 + 1 x + 2 x^{2}+1x+2 since a and b are roots in this polynomial. Thus 10p+q=10*1+2=12.

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Your answer is not clear,please explain briefly

Akhil Bansal - 5 years, 11 months ago

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