Explained: "How to solve x+1=5x+1?"

stampedazt0

Answered question

2023-01-07

How to solve

\sqrt{x} + 1 = \sqrt{5 x + 1}

Elianna Bradley

Beginner2023-01-08Added 8 answers

We have:

\sqrt{x} + 1 = \sqrt{5 x + 1}

Square both sides:

x + 2 \sqrt{x} + 1 = 5 x + 1

Combine like terms:

- 4 x + 2 \sqrt{x} = 0

Divide both sides by -2:

2 x - \sqrt{x} = 0

We can write x as

{(\sqrt{x})}^{2}

2 {(\sqrt{x})}^{2} - \sqrt{x} = 0

Factor:

\sqrt{x} (2 \sqrt{x} - 1) = 0

Two equations result from this:

\sqrt{x} = 0 and 2 \sqrt{x} - 1 = 0

Only 1 equation needs simplification:

\sqrt{x} = 0 and \sqrt{x} = \frac{1}{2}

Square both:

x = 0 and x = \frac{1}{4}

Check:

\sqrt{0} + 1 = \sqrt{5 (0) + 1}

and

\sqrt{\frac{1}{4}} + 1 = \sqrt{5 (\frac{1}{4}) + 1}

You can't use 'macro parameter character #' in math mode

Both roots check.

samkenndcnt

Beginner2023-01-09Added 1 answers

Given:

\sqrt{x} + 1 = \sqrt{5 x + 1}

Square both sides of the equation, taking into account the possibility of additional solutions, to obtain

x + 2 \sqrt{x} + 1 = 5 x + 1

Subtract #x+1# from both sides to get:

2 \sqrt{x} = 4 x

Note that x=0 is a solution. For other solutions, divide both sides by

4 \sqrt{x}

to get:

\frac{1}{2} = \sqrt{x}

Then square both sides and transpose to get:

x = \frac{1}{4}

Check both of our derived solutions:

\sqrt{0} + 1 = 0 + 1 = 1 = \sqrt{1} = \sqrt{0 + 1} = \sqrt{5 (0) + 1}

\sqrt{\frac{1}{4}} + 1 = \frac{1}{2} + 1 = \frac{3}{2} = \sqrt{\frac{9}{4}} = \sqrt{\frac{5}{4} + 1} = \sqrt{5 (\frac{1}{4}) + 1}

So both are solutions of the original equation.

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