Just X X ?

Algebra Level 4

( x 2 5 x + 5 ) x 2 11 x + 30 = 1 \left( { x }^{ 2 }-5x+5 \right) ^{ { x }^{ 2 }-11x+30 }=1

How many possible values of x x satisfy the equation above?


The answer is 6.

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1 solution

Rohit Udaiwal
Jan 24, 2016

Case 1 : x 2 5 x + 5 = 1 x 2 5 x + 4 = 0 x = 5 ± 9 2 x = 1 , 4 x^2-5x+5=1 \implies x^2-5x+4=0 \implies x=\dfrac{5 \pm \sqrt{9}}{2} \implies x=1,4 (By Quadratic Formula .)

Case 2 : x 2 11 x + 30 = 0 x = 11 ± 121 120 2 x = 11 ± 1 2 x = 5 , 6 x^2-11x+30=0 \implies x=\dfrac{11 \pm \sqrt{121-120}}{2} \implies x=\dfrac{11 \pm 1}{2} \implies x=5,6

Cae 3 : x 2 5 x + 5 = 1 x^2-5x+5=-1 and x 2 11 x + 30 x^2-11x+30 is even.

x 2 5 x + 5 = 1 x 2 5 x + 6 = 0 x = 5 ± 25 24 2 x = 5 ± 1 2 x = 2 , 3. x^2-5x+5=-1 \implies x^2-5x+6=0 \implies x=\dfrac{5 \pm \sqrt{25-24}}{2} \implies x=\dfrac{5 \pm1}{2} \\ \implies x=2,3.

And x 2 11 x + 30 x^2-11x+30 is even for its value with 2 2 and 3 3 .Hence x = 1 , 2 , 3 , 4 , 5 , 6. \large x=1,2,3,4,5,6.

Quite a famous variety of problem on brilliant... BTW N i c e Nice ~

Rishabh Jain - 5 years, 4 months ago

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Thank You ! \large \color{#20A900}{\text{Thank You !}}

Rohit Udaiwal - 5 years, 4 months ago

For completeness you might also verify that when x=5,6 you don't have 0^0. You don't in this problem but it's always good to check and be sure you don't have something undefined happening.

Jonathan Hocker - 5 years, 4 months ago

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