It's getting complex

Algebra Level 3

When x = 3 + 5 i 2 \color{#20A900}{x}=\dfrac {3+5i}{2} ,find the value of 2 x 3 + 2 x 2 − 7 x + 72 2\color{#20A900}{x}^3+2\color{#20A900}{x}^2-7\color{#20A900}{x}+72 .

Note : i = − 1 i=\sqrt {-1} .


The answer is 4.

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Rohit Udaiwal by Rohit Udaiwal , 20, India

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3 solutions

Aareyan Manzoor
Dec 14, 2015

x is a root of 2 ( x − 3 + 5 i 2 ) ( x − 3 − 5 i 2 ) = 2 x 2 − 6 x + 17 = 0 2(x-\frac{3+5i}{2})(x-\frac{3-5i}{2})=2x^2-6x+17=0 this is easy to expand because we can just sum and multiply by vietas. now, we mulyiply by (x+4) ( 2 x 2 − 6 x + 17 ) ( x + 4 ) = 2 x 3 + 2 x 2 − 7 x + 68 = 0 (2x^2-6x+17)(x+4)=2x^3+2x^2-7x+68=0 add 4 to both sides 2 x 3 + 2 x 2 − 7 x + 72 = 4 2x^3+2x^2-7x+72=\boxed{4}

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There's a small typo in the first line where you have 2x^2-6x+(1) instead of 2x^2-6x+(17). You have the 17 in the next step just forgot it in that first line. Nice observation about forming that polynomial. What was your process in the construction of that polynomial? The first part using vieta's was clever but the x+4 factor seems pretty arbitrary although it ended up working out very nicely. I ended up just plugging in the value for x lol (I factored out the 2x^2 for my convenience in the calculation).

Jonathan Hocker - 5 years, 6 months ago

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thanks, corrected. we want the term with x 2 x^2 to be 2. it is easy to find x+4 to do that. since the answer is obvously real we can deduce that a -7 will come with the x. we expand to see good luck!

Aareyan Manzoor - 5 years, 6 months ago
Arjen Vreugdenhil
Dec 15, 2015

Since Brilliant's input here requires an integer, it stands to reason that we only need the real part. This is

3 3 − 3 ⋅ 3 ⋅ 5 2 4 + 3 2 − 5 2 2 − 7 ⋅ 3 2 + 72 = − 198 4 + − 16 2 − 21 2 + 72 = 4. \frac{3^3 - 3\cdot 3\cdot 5^2}4 + \frac{3^2-5^2}2 - \frac{7\cdot 3}2 + 72 \\ = \frac{-198}4 + \frac{-16}2 - \frac{21}2 + 72 = 4.

There are more elegant solutions, of course. But they take at least at much time :) Here is my alternative solution that might satisfy the purists among you:

Up to complex conjugation, x x is the unique root of a quadratic equation, and working backward in the famous quadratic formula we have a = 1 , b = 3 , b 2 − 4 a c = − 25 ∴ 4 a c = − 34 ∴ c = 8 1 2 , a = 1, b = 3, b^2 - 4ac = -25 \therefore 4ac = -34 \therefore c = 8\tfrac12, so that the equation is x 2 − 3 x + 8 1 2 = 0 ∴ 2 x 2 − 6 x + 17 = 0. x^2 - 3x + 8\tfrac12 = 0\ \therefore\ 2x^2 - 6x + 17 = 0. We can reduce the order of the given polynomial by subtracting out multiples of this: 2 x 3 + 2 x 2 − 7 x + 72 = ( x + 4 ) ( 2 x 2 − 6 x + 17 ) + 4 = 0 + 4 = 4 . 2x^3 + 2x^2 - 7x + 72 = (x + 4)(2x^2 - 6x + 17) + 4 = 0 + 4 = \boxed{4}.

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Andrew Wu
Dec 15, 2015

To me, the most obvious start is to try to factor the polynomial. It factors as ( 2 x 2 − 7 ) ( x + 1 ) + 79 (2x^2 - 7)(x+1) + 79 . Surprisingly enough, you get nice numbers when you expand - we end up with ( 15 ) ( 5 / 2 ) ( i + 1 ) ( i − 1 ) + 79 (15)(5/2)(i+1)(i-1) + 79 after plugging and chugging, which gives us 2 x 3 + 2 x 2 − 7 x + 72 = 4 2x^3 + 2x^2 - 7x + 72 = 4 .

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