Some of Digits - 2

Algebra Level 5

Find the sum of all real values of x x which satisfy:

1 x 2 − 10 x − 45 + 1 x 2 − 10 x − 29 = 2 x 2 − 10 x − 69 \frac{1}{x^{2} - 10x - 45} + \frac{1}{x^{2} -10x - 29} = \frac{2}{x^{2} - 10x - 69}

Try also Some of Digits -1 .


The answer is 10.

Razing Thunder by Razing Thunder , 16, India

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2 solutions

Chew-Seong Cheong
Jul 10, 2020

1 x 2 − 10 x − 45 + 1 x 2 − 10 x − 29 = 2 x 2 − 10 x − 69 Let u = x 2 − 10 x − 37 1 u − 8 + 1 u + 8 = 2 u − 32 Multiply both sides by ( u − 8 ) ( u + 8 ) u + 8 + u − 8 = 2 ( u 2 − 64 ) u − 32 u 2 − 32 u = u 2 − 64 ⟹ u = 2 x 2 − 10 x − 37 = 2 x 2 − 10 x − 39 = 0 ( x + 3 ) ( x − 13 ) = 0 ⟹ x = − 3 , 13 \begin{aligned} \frac 1{x^2-10x-45} + \frac 1{x^2-10x-29} & = \frac 2{x^2-10x-69} & \small \blue{\text{Let }u = x^2-10x-37} \\ \frac 1{u-8} + \frac 1{u+8} & = \frac 2{u-32} & \small \blue{\text{Multiply both sides by }(u-8)(u+8)} \\ u+8 + u - 8 & = \frac {2(u^2-64)}{u-32} \\ u^2 - 32u & = u^2 - 64 \\ \implies u & = 2 \\ x^2 - 10x - 37 & = 2 \\ x^2 - 10x - 39 & = 0 \\ (x+3)(x-13) & = 0 \\ \implies x & = -3, 13 \end{aligned}

Therefore the sum of roots is − 3 + 13 = 10 -3+13 = \boxed{10} .

2 Helpful 1 Interesting 3 Brilliant 0 Confused
Sahil Goyat
Jul 9, 2020

0 Helpful 0 Interesting 0 Brilliant 2 Confused

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