Tweaking radicals

Algebra Level 2

2 − x − 4 − 3 x 2 − x + 4 − 3 x = 6 − 5 x − 8 − 7 x 6 − 5 x + 8 − 7 x , x = ? \large \dfrac{\sqrt{2-x} - \sqrt{4-3x}}{\sqrt{2-x} + \sqrt{4-3x}} = \dfrac{\sqrt{6-5x} - \sqrt{8-7x}}{\sqrt{6-5x} + \sqrt{8-7x}} \quad,\quad x = \ ?

3 2 4 1

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Chung Kevin by Chung Kevin , 45, Australia

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2 solutions

Rohit Udaiwal
Sep 27, 2015

Use componendo-dividendo to get : 2 − x 4 − 3 x = 6 − 5 x 8 − 7 x Square both sides and cross multiply ( 4 − 3 x ) ( 6 − 5 x ) = ( 2 − x ) ( 8 − 7 x ) ⇒ 24 − 20 x − 18 x + 15 x 2 = 16 − 14 x − 8 x + 7 x 2 ⇒ 24 − 38 x + 15 x 2 − 16 + 22 x − 7 x 2 = 0 ⇒ 8 x 2 − 16 x + 8 = 0 ⇒ 8 ( x 2 − 2 x + 1 ) = 0 ⇒ x 2 − 2 x + 1 = 0 ⇒ ( x − 1 ) 2 = 0 ⇒ x = 1 \color{#3D99F6}{\sqrt {\frac {2-x}{4-3x}}=\sqrt {\frac {6-5x}{8-7x}}} \\ \text {Square both sides and cross multiply} \\ (4-3x)(6-5x)=(2-x)(8-7x) \\ \Rightarrow{24-20x-18x+15x^{2}=16-14x-8x+7x^{2}} \\ \Rightarrow{24-38x+15x^{2}-16+22x-7x^{2}=0} \\ \Rightarrow{8x^{2}-16x+8=0} \\ \Rightarrow{8 (x^{2}-2x+1)=0} \\ \Rightarrow{x^{2}-2x+1=0} \\ \Rightarrow{(x-1)^{2}=0} \\ \color{#20A900}{\Rightarrow{x=1}}

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Sam Maltia
Oct 18, 2015

All the rest of the solutions in one way or another give you an imaginary solution; therefore, the answer has to be 1.

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