Tukukkk

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?

You made a mistake when completing the square.

$$x^2-frac{1}{4}x = frac{3}{4} color{red}{impliesbig(x-frac{1}{2}big)^2 = 1}$$

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

Note that the equation is rewritten such that $a = 1$, so you need to add $big(frac{b}{2}big)^2$ 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.

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

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

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 $2x = pmsqrt{x+3}$. 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.

It lies with how you chose to complete the square. We begin at this line:

$$x^2 - frac14 x - frac34 = 0$$

Recall that $(x - b)^2 = x^2 - 2bx + b^2$. In this case, then, $2bx = frac14x$, which, solving for $b$, means $b = 1/8$.

Thus, we want a term of $(x - 1/8)^2$, which is equal to $x^2 - frac14x + frac{1}{64}$.

Then, we have

$$begin{align*}
x^2 - frac14 x - frac34 &= x^2 - frac14 x - frac34 + frac{1}{64} - frac{1}{64} \
&= left(x^2 - frac14 x + frac{1}{64}right) - frac34 - frac{1}{64} \
&= left(x - frac18 right)^2 - frac{49}{64} = 0
end{align*}$$

Thus, our solution can be found by solving

$$left(x - frac18 right)^2 - frac{49}{64} = 0$$

Don't forget to double-check the solutions by plugging them into the original equation in case an extraneous solution was introduced.