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Algebra Level 5

Give the sum!(round to 2 decimal places)


The answer is -3.67.

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Mardokay Mosazghi by Mardokay Mosazghi , 22, USA

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2 solutions

Chew-Seong Cheong
Oct 12, 2014

4 x 2 + 15 x + 17 x 2 + 4 x + 12 = 5 x 2 + 16 x + 18 2 x 2 + 5 x + 13 \dfrac {4x^2+15x+17} {x^2+4x+12} = \dfrac {5x^2+16x+18} {2x^2+5x+13}

⇒ 4 x 2 + 15 x + 17 x 2 + 4 x + 12 = 4 x 2 + 15 x + 17 + x 2 + x + 1 x 2 + 4 x + 12 + x 2 + x + 1 \Rightarrow \dfrac {4x^2+15x+17} {x^2+4x+12} = \dfrac {4x^2+15x+17 + x^2 + x + 1} {x^2+4x+12 + x^2+x+1}

( 4 x 2 + 15 x + 17 ) ( x 2 + 4 x + 12 ) + ( 4 x 2 + 15 x + 17 ) ( x 2 + x + 1 ) = ( 4 x 2 + 15 x + 17 ) ( x 2 + 4 x + 12 ) + ( x 2 + 4 x + 12 ) ( x 2 + x + 1 ) (4x^2+15x+17) (x^2+4x+12)+ (4x^2+15x+17)(x^2+x+1) = (4x^2+15x+17)(x^2+4x+12) + (x^2+4x+12)(x^2+x+1)

( 4 x 2 + 15 x + 17 ) ( x 2 + x + 1 ) = ( x 2 + 4 x + 12 ) ( x 2 + x + 1 ) (4x^2+15x+17)(x^2+x+1) = (x^2+4x+12)(x^2+x+1)

4 x 2 + 15 x + 17 = x 2 + 4 x + 12 ⇒ 3 x 2 + 11 x + 5 = 0 4x^2+15x+17 = x^2+4x+12 \quad \Rightarrow 3x^2+11x+5 = 0

⇒ x = − 11 ± 121 − 60 6 = − 11 ± 61 6 \Rightarrow x = \dfrac {-11 \pm \sqrt{121-60} }{6} = \dfrac {-11 \pm \sqrt{61} }{6}

The required solution = − 11 + 61 6 + − 11 − 61 6 = − 11 3 ≈ − 3.67 \frac {-11 + \sqrt{61} }{6} + \frac {-11 - \sqrt{61} }{6} = \frac{-11}{3} \approx \boxed {-3.67}

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@Chew-Seong Cheong Sir nice solution, i was thinking one with vieta formula but nice solution!!

Mardokay Mosazghi - 6 years, 8 months ago

You can't just ignore the (x^2+x+1)!

Kenny Lau - 6 years, 7 months ago

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There are no real solutions to x 2 + x + 1 = 0 x^2 + x + 1 = 0

Calvin Lin Staff - 6 years, 7 months ago
Akul Agrawal
Nov 2, 2015

Just applied componendo-dividendo

4 x 2 + 15 x + 17 − ( x 2 + 4 x + 12 ) 4 x 2 + 15 x + 17 + ( x 2 + 4 x + 12 ) = 5 x 2 + 16 x + 18 − ( 2 x 2 + 5 x + 18 ) 5 x 2 + 16 x + 18 + ( 2 x 2 + 5 x + 18 ) ⇒ 3 x 2 + 11 x + 5 5 x 2 + 19 x + 29 = 3 x 2 + 11 x + 5 7 x 2 + 21 x + 31 \frac { 4{ x }^{ 2 }+15x+17-({ x }^{ 2 }+4x+12) }{ 4{ x }^{ 2 }+15x+17+({ x }^{ 2 }+4x+12) } =\frac { 5{ x }^{ 2 }+16x+18-(2{ x }^{ 2 }+5x+18) }{ 5{ x }^{ 2 }+16x+18+(2{ x }^{ 2 }+5x+18) } \\ \Rightarrow \frac { 3{ x }^{ 2 }+11x+5 }{ 5{ x }^{ 2 }+19x+29 } =\frac { 3{ x }^{ 2 }+11x+5 }{ 7{ x }^{ 2 }+21x+31 }

Now the factors are easily visible.

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