Radical Equation ~ 1

Algebra Level 2

Find the sum of all x x satisfying the equation below.

3 x = 1 + 3 x − 1 1 + x + 1 \sqrt{3x}=1+\frac{3x-1}{1+\sqrt{x+1}}

1 3 \frac{1}{3} 5 6 \frac{5}{6} 5 5 No Solution 3 3

Achmad Damanhuri by Achmad Damanhuri , 26, Indonesia

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1 solution

Chew-Seong Cheong
Mar 11, 2019

3 x = 1 + 3 x − 1 1 + x + 1 3 x − 1 − ( 3 x − 1 ) ( 3 x + 1 ) 1 + x + 1 = 0 ( 3 x − 1 ) ( 1 − 3 x + 1 1 + x + 1 ) = 0 \begin{aligned} \sqrt{3x} & = 1 + \frac {\color{#3D99F6}3x-1}{1+\sqrt{x+1}} \\ \sqrt{3x} - 1 - \frac {\color{#3D99F6}(\sqrt{3x}-1)(\sqrt{3x}+1)}{1+\sqrt{x+1}} & = 0 \\ \left(\sqrt{3x} - 1\right)\left(1 - \frac {\sqrt{3x}+1}{1+\sqrt{x+1}} \right) & = 0 \end{aligned}

⟹ { 3 x − 1 = 0 ⟹ 3 x = 1 ⟹ x = 1 3 1 − 3 x + 1 1 + x + 1 = 0 ⟹ 3 x + 1 = 1 + x + 1 ⟹ x = 1 2 \implies \begin{cases} \sqrt{3x} - 1 = 0 & \implies 3x = 1 & \implies x = \dfrac 13 \\ 1 - \dfrac {\sqrt{3x}+1}{1+\sqrt{x+1}} = 0 & \implies \sqrt{3x}+1 = 1+\sqrt{x+1} & \implies x = \dfrac 12 \end{cases}

Therefore, the sum of all x x satisfying the equation is 1 3 + 1 2 = 5 6 \dfrac 13+\dfrac 12 = \boxed{\dfrac 56} .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

@Achmad Damanhuri , it is better to use a proper sentence instead ending the sentence hanging with "is...". You can use "Find the sum of all x x satisfying the equation below." to end as a sentence OR "What is the sum of all x x satisfying the equation below?" to end as a question. The word "radical" may not be necessary.

Chew-Seong Cheong - 2 years, 3 months ago

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