Double square
⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎧ a = 4 − 5 − a b = 4 + 5 − b c = 4 − 5 + c d = 4 + 5 + d
Find the product a b c d .
Hint:
You don't need to find any of the individual variables!
The answer is 11.
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2 solutions
Moderator note:
The statement "the product of the four roots is the constant term of the polynomial" is based off of Vieta's Formula and is true in particular here because the degree of the polynomial is even and leading coefficient is 1. In general, if n is the degree of a polynomial, the leading coefficient is a , and the constant term is z , the product of the roots in the polynomial is ( − 1 ) n a z .
Very clever!
Great! That's a nice way to explain the pairing of roots in Paola's solution.
Doesn't it remain to check that such real numbers indeed exist?
Raising to square twice each equality is got it:
a 4 − 8 a 2 + a + 1 1 = 0
b 4 − 8 b 2 + b + 1 1 = 0
c 4 − 8 c 2 − c + 1 1 = 0
d 4 − 8 d 2 − d + 1 1 = 0
Raising again to square each of this equalities is obtained expressions like x 8 − 1 6 x 6 + 8 6 x 4 − 1 7 7 x 2 + 1 2 1 = 0 for a , b , c and d . Let be y = x 2 , the last expression is y 4 + 1 6 y 3 + 8 6 y 2 − 1 7 7 y + 1 2 1 = 0 this expression has four complex roots y 1 , y 2 , y 3 and y 4 so that y 1 y 2 y 3 y 4 = 1 2 1 . As a 2 b 2 c 2 d 2 = y 1 y 2 y 3 y 4 ⇒ a b c d = 1 2 1 = 1 1
The 8th degree equation has 8 roots, and you don't know which 4 of the 8 roots your a, b, c, d correspond to. So you can't be sure that abcd = 11.
In other words, the 8 roots multiply to 121. But we're not sure if 4 of them multiply to 11.
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Indeed. There needs to be some justification of how the roots pair up (e.g. see my comment below), in order to conclude that we have the 4 that multiply to 11.
Can you elaborate on "Raising again to square each of this equalities is obtained expressions"?
IE ( a 4 − 8 a 2 + a + 1 1 ) 2 does not have that given form.
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Let me check it again
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What is actually happening, is that the 4 variables are positive numbers that satisfy x 2 − 4 = ± 5 ± x .
By squaring this, we get ( x 2 − 4 ) 2 = 5 ± x ⇒ x 4 − 8 x 2 + 1 1 = ± x . Now, squaring both sides again, we get ( x 4 − 8 x 2 + 1 1 ) 2 = x 2 . This is the final expression x 8 − 1 6 x 6 + 8 6 x 4 − 1 7 7 x 2 + 1 2 1 = 0 that you get.
By repeated squaring, we ended up introducing extraneous roots, and have to account for them.
Starting along the lines of Paola's soluion: Squaring and re-arranging, and re-squaring the four equations gives us:
a 4 − 8 a 2 + a + 1 1 = 0
b 4 − 8 b 2 + b + 1 1 = 0
c 4 − 8 c 2 − c + 1 1 = 0
d 4 − 8 d 2 − d + 1 1 = 0
Thus the four roots of f ( x ) = x 4 − 8 x 2 + x + 1 1 = 0 are the values a , b , − c , and − d . The product of the four roots is the constant term of the polynomial, i.e.
1 1 = a b ( − c ) ( − d ) = a b c d