Mistake in solving an equation involving a square root

Question

I want to solve $2x = \sqrt{x+3}$, which I have tried as below:

$$\begin{equation} 4x^2 - x -3 = 0 \\ x^2 - \frac14 x - \frac34 = 0 \\ x^2 - \frac14x = \frac34 \\ \left(x - \frac12 \right)^2 = 1 \\ x = \frac32 , -\frac12 \end{equation}$$

This, however, is incorrect.

What is wrong with my solution?

+1 for showing your full working. Keep up the good work and I hope my answer is satisfactory! — Hugh Entwistle, Dec 26 '18 at 12:23

KM101 · Accepted Answer · 2018-12-27T06:41:26.200

You made a mistake when completing the square.

$$x^2-\frac{1}{4}x = \frac{3}{4} \color{red}{\implies\left(x-\frac{1}{2}\right)^2 = 1}$$

This is easy to spot since $(a\pm b)^2 = a^2\pm2ab+b^2$, which means the coefficient of the linear term becomes $-2\left(\frac{1}{2}\right) = -1 \color{red}{\neq -\frac{1}{4}}$. This means something isn’t correct...

Note that the equation is rewritten such that $a = 1$, so you need to add $\left(\frac{b}{2}\right)^2$ to both sides and factor. (In other words, divide the coefficient of the linear term $x$ by $2$ and square the result, which will then be added to both sides.)

$$b = -\frac{1}{4} \implies \left(\frac{b}{2}\right)^2 \implies \frac{1}{64}$$

Which gets

$$x^2-\frac{1}{4}x+\color{blue}{\frac{1}{64}} = \frac{3}{4}+\color{blue}{\frac{1}{64}}$$

Factoring the perfect square trinomial yields

$$\left(x-\frac{1}{8}\right)^2 = \frac{49}{64}$$

And you can probably take it on from here.

Edit: As it has been noted in the other answers (should have clarified this as well), squaring introduces the possibility of extraneous solutions, so always check your solutions by plugging in the values obtained in the original equation. By squaring, you’re solving

$$4x^2 = x+3$$

which is actually

$$2x = \color{blue}{\pm}\sqrt{x+3}$$

so your negative solution will satisfy this new equation but not the original one, since that one is

$$2x = \sqrt{x+3}$$

with no $\pm$.

Hugh Entwistle · Answer 2 · 2018-12-30T03:39:58.037

11

From

$$x^2-\frac{1}{4}x=\frac{3}{4}$$ to $$\left(x-\frac{1}{2} \right)^2=1$$ you have not completed the square correctly.

It should instead be

$$\left(x-\frac{1}{8} \right)^2-\frac{1}{64}=\frac{3}{4}$$

Additionally please note that, if we were to proceed with your original solution as 'correct', one of your solutions does in fact not work. When we originally square both sides we have, in a sense, 'modified' the question -- allowing for negative solutions for $x$. Your original solution of $x=-\frac{1}{2}$ does not satisfy the original equation - we must be mindful to check our solutions in these situations.

edited Dec 30 '18 at 03:39

answered Dec 26 '18 at 08:37

Hugh Entwistle

1,239

4

The OP's answer was also incorrect because one of his/her solutions was extraneous; it might be a good idea to add a warning about that to your answer as well. – Franklin Pezzuti Dyer Dec 26 '18 at 22:22
@Frpzzd I have edited my answer to reflect your concern! – Hugh Entwistle Dec 30 '18 at 03:40

score 7 · Answer 3 · answered Dec 26 '18 at 08:44

7

Two mistakes:

1) mistake in completing the square. Remember, divide the coefficient of the $x$ term by two, not multiply. You should've got: $(x-\frac 18)^2 = \frac 34 + \frac{1}{64}$.

2) when you square, you run the risk of introducing "redundant roots". This is because when you solve by squaring, you're really solving $2x = \pm\sqrt{x+3}$. So put your solutions back into the original to see which one(s) satisfy the original equation, and discard the rest. This applies whenever you raise both sides of an equation to any even power.

answered Dec 26 '18 at 08:44

Deepak

26,801

3

This is the only completely correct answer. – daviewales Dec 27 '18 at 00:14
Yep, this answer is concise, complete, and accurate. Wish that it was the accepted one. – JonathanZ Dec 27 '18 at 16:57

score 2 · Answer 4 · answered Dec 26 '18 at 12:55

2

$$\begin{equation} 4x^2 - x -3 = 0 \\ (4x + 3)(x - 1) = 0 \\ x = -\frac34 , 1 \end{equation}$$

answered Dec 26 '18 at 12:55

Leonhard Euler

631

Upvoting, although it would be nice to see a little explanation given the level of the asker. – user1717828 Dec 26 '18 at 14:37
2

How can $x$ be negative if $2x$ is equal to a square root? – Jon Dec 26 '18 at 17:44
@Jon Square roots have $2$ solutions, a positive and a negative. – Leonhard Euler Dec 26 '18 at 21:49
These are the two solutions of the quadratic equation $4x^2-x-3=0$. Only one of these is the solution to $2x = \sqrt{x+3}$. See this question. – daviewales Dec 27 '18 at 00:13
Before you can claim $4x^2 - x - 3 = 0$, question is given as $2x = \sqrt{x+3}$. This requires that $x \ge 0$. The radical sign distinctly refers only to the positive root. When we square both sides to get $4x^2 = x+3$ this allows for the extraneous solution to $2x = -\sqrt{x+3}$. We must make note that $x \ge 0$. Thus $(4x+3)(x-1) =0$ and as $x \ge 0$ then $4x+3 \ge 3$ so $4x+3\ne 0$ so $x-1 = 0$ and $x = 1$. – fleablood Dec 27 '18 at 07:25
@fleablood I'm afraid that this is a cultural (presumably North American) thing. In the UK, where I did my degree in Maths, a square root is a square root and has $2$ solutions. – Leonhard Euler Dec 27 '18 at 12:16
Actually, it’s by convention. The use of a square root with no sign in front indicates the principal square root, which is a non-negative value. Hence, $x = \sqrt a$ shouldn’t be confused for $x^2 = a \iff x = \color{blue}{\pm} \sqrt a$. – KM101 Dec 27 '18 at 13:33
1

'In the UK, where I did my degree in Maths, a square root is a square root and has 2 solutions.' Could you provide a reference for this. I highly doubt that this is generally true. (What's sometimes done, though mostly not in the context relevant here, is that the symbol denotes one unspecified root. Yet it won't denote both.) – quid Dec 27 '18 at 15:36
@quid Why don't you provide a reference that the square root is always positive, in the UK as well, and then I'll find a reference for you to the contrary? (Doesn't my degree suffice?) – Leonhard Euler Dec 27 '18 at 20:13
@KM101 "by convention" where you live perhaps. In the UK, a square root has $2$ solutions. – Leonhard Euler Dec 27 '18 at 20:15
Exercise 1 there makes no sense unless the sign of the root is fixed. – quid Dec 27 '18 at 20:17
You say "a square root has $2$ solutions", so what mathematical object is $\sqrt{5}$ exactly? And what is $\sqrt{3} + \sqrt{5}$; does this have four solutions then? – quid Dec 27 '18 at 20:24
@quid It depends on the context of the question. In this question, I see no reason why $-\frac34$ cannot be correct. – Leonhard Euler Dec 27 '18 at 20:25
Let us continue this discussion in chat. – Leonhard Euler Dec 27 '18 at 20:26

fleablood · Answer 5 · 2018-12-27T07:27:19.117

$\require{cancel}$Additional details in $\color{blue}{blue}$. Important detail in $\color{green}{green}$. Mistakes in $\color{red}{\cancel{\text{canceled red}}}$. Corrections in $\color{purple}{purple}$.

$\begin{equation} \color{blue}{2x = \sqrt{x+3}}\color{green}{\ge 0}\\ \color{blue}{4x^2 = x+3}\color{green}{\text{!AND!} x\ge 0}\\ 4x^2 - x -3 = 0 \\ x^2 - \frac14 x - \frac34 = 0 \\ x^2 - \frac14x = \frac34 \\ \color{blue}{x^2 -2\cdot\frac 18=\frac 34}\\ \color{blue}{x^2 -2\cdot\frac 18 +(\frac 18)^2=\frac 34+\frac 1{64}}\\ \color{red}{\cancel{(x - \frac12 )^2 = 1}}\color{purple}{(x - \frac18)^2 = \frac {49}{64}} \\ \color{blue}{x-\frac 18=\pm \frac 78}\\ \color{red}{\cancel{x = \frac32 , -\frac12}}\color{purple}{x = 1 , -\frac34}\color{green}{\text{!AND!} x\ge 0}\\ \color{purple}{x = 1}\\ \end{equation}$

Mistake in solving an equation involving a square root

5 Answers5