Fraction

Algebra Level 3

It is given that x + 1 x = 4 x+\dfrac{1}{x}=4 and A x 2 x 4 + x 2 + 1 = 49 750 \dfrac{Ax^{2}}{x^{4}+x^{2}+1}=\dfrac{49}{750} .

If the value of A A can be expressed as m n \dfrac {m}{n} , where m m and n n are coprime positive integers, what is the value of m + n m+n ?


The answer is 99.

I Gede Arya Raditya Parameswara by I Gede Arya Raditya Para… , 17

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2 solutions

A x 2 x 4 + x 2 + 1 = 49 750 Dividing LHS up and down by x 2 A x 2 + 1 + 1 x 2 = 49 750 A x 2 + 2 + 1 x 2 − 1 = 49 750 A ( x + 1 x ) 2 − 1 = 49 750 Note that x + 1 x = 4 A 16 − 1 = 49 750 A 15 = 49 750 A = 49 × 15 750 = 49 50 \begin{aligned} \frac {Ax^2}{x^4+x^2+1} & = \frac{49}{750} & \small \color{#3D99F6} \text{Dividing LHS up and down by }x^2 \\ \frac A {x^2+1+\dfrac 1{x^2}} & = \frac{49}{750} \\ \frac A {{\color{#3D99F6}x^2+2+\dfrac 1{x^2}}-1} & = \frac{49}{750} \\ \frac A {{\color{#3D99F6}\left(x+\dfrac 1x\right)^2}-1} & = \frac{49}{750} & \small \color{#3D99F6} \text{Note that }x+\frac 1x = 4 \\ \frac A {{\color{#3D99F6}16}-1 } & = \frac{49}{750} \\ \frac A {15} & = \frac{49}{750} \\ A & = \frac{49 \times 15}{750} \\ & = \frac {49}{50} \end{aligned}

⟹ m + n = 49 + 50 = 99 \implies m + n = 49+50 = \boxed{99}

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x + 1 x = 4 x+\frac{1}{x}=4 Squaring both sides, x 2 + 1 x 2 = 14 x^2+\frac{1}{x^2}=14 x 4 + 1 = 14 x 2 x^4+1=14x^2 Therefore, A x 2 ( x 4 + 1 ) + x 2 = A x 2 14 x 2 + x 2 = A 15 \frac{Ax^2}{(x^4+1)+x^2} =\frac{Ax^2}{14x^2+x^2} =\frac{A}{15} So, A 15 = 49 750 \frac{A}{15}=\frac{49}{750} A = 49 50 A=\frac{49}{50} ⟹ m + n = 99 \implies m+n=\boxed{99}

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