A Crazy Quartic

Level pending

Let a , b , c , d a,b,c,d be the roots of the following equation.

x 4 + 10 x 3 + k x 2 + 100 x − 1001 = 0 x^4+10x^3+kx^2+100x-1001=0

If a b = 77 ab=77 , the value of k k can be expressed as a mixed number p q r p\frac{q}{r} , where p p is a positive integer, and q q and r r are coprime positive integers. Find the value of p + q + r p+q+r .


The answer is 166.

Ryan Phua by Ryan Phua , 22, Singapore

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1 solution

Sujoy Roy
Dec 17, 2013

a + b + c + d = − 10 , a b + a c + a d + b c + b d + c d = k , a b c + b c d + c d a + d a b = − 100 a+b+c+d=-10, ab+ac+ad+bc+bd+cd=k, abc+bcd+cda+dab=-100 and a b c d = − 1001. abcd=-1001.

As a b = 77 ab=77 , c d = − 13 cd=-13 . From a b c + b c d + c d a + d a b = − 100 abc+bcd+cda+dab=-100 we get, a b ( c + d ) + c d ( a + b ) = − 100 \ ⇒ 77 ( c + d ) − 13 ( a + b ) = − 100 \ ⇒ 77 ( c + d ) + 13 ( 10 + c + d ) = − 100 \ ⇒ c + d = − 23 9 ab(c+d)+cd(a+b)=-100 \ \Rightarrow 77(c+d)-13(a+b)=-100 \ \Rightarrow 77(c+d) +13(10+c+d)=-100 \ \Rightarrow c+d=-\frac{23}{9}

So, a + b = − 67 9 a+b=-\frac{67}{9}

Now, k = a b + c d + ( a + b ) ( c + d ) = 77 − 13 + − 23 9 ∗ − 67 9 = 83 2 81 k=ab+cd+(a+b)(c+d)=77-13+\frac{-23}{9}*\frac{-67}{9}=83\frac{2}{81} . So, p + q + r = 83 + 2 + 81 = 166 p+q+r=83+2+81=\boxed{166}

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