Vieta's Theorem is a good approach!

Algebra Level 2

Find three real numbers a < b < c a < b < c satisfying:

a + b + c = a + b + c = 21 4 \frac{21}{4}

1 / a + 1 / b + 1 / c = 1/a + 1/b + 1/c = 21 4 \frac{21}{4}

a b c = 1. abc = 1.

(4, 1, 1/4) (i, -i, 1) (1/4, 1, 4) (1, 4, 1/4)

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2 solutions

Mufri Gani
Jun 30, 2018

Relevant wiki: Vieta's Formula Problem Solving - Intermediate

a + b + c = a + b + c = 21 4 \frac{21}{4}

1 / a + 1 / b + 1 / c = 1/a + 1/b + 1/c = 21 4 \frac{21}{4}

a b c = 1 abc = 1

1 / a + 1 / b + 1 / c = 21 / 4 1/a + 1/b + 1/c = 21/4

( b c + a c + a b ) / ( a b c ) = 21 / 4 (bc+ac+ab)/(abc) = 21/4 )

( b c + a c + a b ) / 1 = 21 / 4 (bc+ac+ab)/1 = 21/4

b c + a c + a b = 21 / 4 bc + ac + ab = 21/4

Let a, b, c be the roots of a certain 3-degree polynomial equation. It follows that…

( x ³ ( a + b + c ) x ² + ( a b + a c + b c ) x ( a b c ) = 0 (x³ - (a+b+c)x² + (ab+ac+bc)x - (abc) = 0

x ³ ( 21 / 4 ) x ² + ( 21 / 4 ) x 1 = 0 x³ - (21/4)x² + (21/4)x - 1 = 0

4 x ³ 21 x ² + 21 x 4 = 0 4x³ - 21x² + 21x - 4 = 0

( 4 x ³ 4 ) + ( 21 x ² + 21 x ) = 0 (4x³ - 4) + (-21x² + 21x) = 0

4 ( x ³ 1 ) + ( 21 x ) ( x 1 ) = 0 4(x³ - 1) + (-21x)(x - 1) = 0

4 ( x 1 ) ( x ² + x + 1 ) 21 x ( x 1 ) = 0 4(x - 1)(x² + x + 1) - 21x(x - 1) = 0

( x 1 ) [ 4 ( x ² + x + 1 ) 21 x ] = 0 (x - 1)[4(x² + x + 1) - 21x] = 0

( x 1 ) ( 4 x ² + 4 x + 4 21 x ) = 0 (x - 1)(4x² + 4x + 4 - 21x) = 0

( x 1 ) ( 4 x ² 17 x + 4 ) = 0 (x - 1)(4x² - 17x + 4) = 0

( x 1 ) ( 4 x 1 ) ( x 4 ) = 0 (x - 1)(4x - 1)(x - 4) = 0

x = 1 , x = 1 / 4 , x = 4. x = 1, x = 1/4, x = 4.

Since a < b < c a < b < c , therefore…

a = 1 4 a = \frac{1}{4}

b = 1 b = 1

c = 4 c = 4

When you mentioned a < b < c a<b<c , every one can guess the solution.

Chew-Seong Cheong - 2 years, 11 months ago
Aaryan Vaishya
Aug 9, 2019

Just test the first choice and find that the second choice has the same values organized correctly.

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