Exponent variable

Algebra Level 3

( 5 + 24 ) x 2 − 3 + ( 5 − 24 ) x 2 − 3 = 10 , x = ? \large (5 + \sqrt {24})^{x^2-3} + (5-\sqrt{24})^{x^2-3} =10, \ \ \ \ \ x = \ ?

{ 2 , − 2 , 2 , − 2 } \{2,-2, \sqrt 2, - \sqrt 2 \} { 2 , 2 } \{2,\sqrt 2 \} None of these choices { 2 , − 2 } \{ \sqrt 2, - \sqrt 2 \} Lack of relevant information { 2 , − 2 } \{ 2, -2 \}

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Harshi Singh by Harshi Singh , 20, India

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1 solution

Chew-Seong Cheong
Apr 26, 2015

( 5 + 24 ) x 2 − 3 + ( 5 − 24 ) x 2 − 3 = 10 ( 25 + 24 ) x 2 − 3 + ( 25 − 24 ) x 2 − 3 = 10 ( 25 + 24 ) 2 ( x 2 − 3 ) + ( 25 + 24 ) x 2 − 3 ( 25 − 24 ) x 2 − 3 = 10 ( 25 + 24 ) x 2 − 3 ( 25 + 24 ) 2 ( x 2 − 3 ) + ( 25 − 24 ) x 2 − 3 = 10 ( 25 + 24 ) x 2 − 3 ( 5 + 24 ) 2 ( x 2 − 3 ) − 10 ( 5 + 24 ) x 2 − 3 + 1 = 0 ⇒ ( 5 + 24 ) x 2 − 3 = 10 ± 100 − 4 2 = 5 ± 24 ⇒ x 2 − 3 = ± 1 ⇒ { x 2 = 4 ⇒ x = ± 2 x 2 = 2 ⇒ x = ± 2 (5+\sqrt{24})^{x^2-3} + (5-\sqrt{24})^{x^2-3} = 10 \\ (\sqrt{25}+\sqrt{24})^{x^2-3} + (\sqrt{25}-\sqrt{24})^{x^2-3} = 10 \\ (\sqrt{25}+\sqrt{24})^{2(x^2-3)} + (\sqrt{25}+\sqrt{24})^{x^2-3}(\sqrt{25}-\sqrt{24})^{x^2-3} = 10 (\sqrt{25}+\sqrt{24})^{x^2-3} \\ (\sqrt{25}+\sqrt{24})^ {2(x^2-3)} + (25-24)^{x^2-3} = 10 (\sqrt{25}+\sqrt{24})^{x^2-3} \\ (5+\sqrt{24})^ {2(x^2-3)} - 10 (5+\sqrt{24})^{x^2-3} + 1 = 0 \\ \Rightarrow (5+\sqrt{24})^{x^2-3} = \dfrac {10 \pm \sqrt{100-4}} {2} = 5 \pm \sqrt{24} \\ \Rightarrow x^2 - 3 = \pm 1 \quad \Rightarrow \begin{cases} x^2 = 4 & \Rightarrow x = \boxed{\pm 2} \\ x^2 = 2 & \Rightarrow x = \boxed{\pm \sqrt{2}} \end{cases}

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Moderator note:

There's still a simpler approach.

@Chew-Seong Cheong , can you please elaborate on your solution, Thank you.

Brilliant Mathematics Staff - 6 years, 1 month ago

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