mjhjmtu

I want to solve $2x = sqrtx+3$, which I have tried as below:

$$beginequation
4x^2 - x -3 = 0 \
x^2 - frac14 x - frac34 = 0 \
x^2 - frac14x = frac34 \
left(x - frac12 right)^2 = 1 \
x = frac32 , -frac12
endequation$$

This, however, is incorrect.

What is wrong with my solution?

You made a mistake when completing the square.

$$x^2-frac14x = frac34 colorredimpliesleft(x-frac12right)^2 = 1$$

This is easy to spot since $(apm b)^2 = a^2pm2ab+b^2$, which means the coefficient of the linear term becomes $-2left(frac12right) = -1 colorredneq -frac14$. This means something isn’t correct...

Note that the equation is rewritten such that $a = 1$, so you need to add $left(fracb2right)^2$ 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.)

$$b = -frac14 implies left(fracb2right)^2 implies frac164$$

Which gets

$$x^2-frac14x+colorbluefrac164 = frac34+colorbluefrac164$$

Factoring the perfect square trinomial yields

$$left(x-frac18right)^2 = frac4964$$

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

$$2x = colorbluepmsqrtx+3$$

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

$$2x = sqrtx+3$$

with no $pm$.

From

$$x^2-frac14x=frac34$$ to $$left(x-frac12 right)^2=1$$ you have not completed the square correctly.

It should instead be

$$left(x-frac18 right)^2-frac164=frac34$$

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 $x=-frac12$ does not satisfy the original equation - we must be mindful to check our solutions in these situations.

Two mistakes:

1) mistake in completing the square. Remember, divide the coefficient of the $x$ term by two, not multiply. You should've got: $(x-frac 18)^2 = frac 34 + frac164$.

2) when you square, you run the risk of introducing "redundant roots". This is because when you solve by squaring, you're really solving $2x = pmsqrtx+3$. 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.

$$beginequation
4x^2 - x -3 = 0 \
(4x + 3)(x - 1) = 0 \
x = -frac34 , 1
endequation$$

$requirecancel$Additional details in $colorblueblue$. Important detail in $colorgreengreen$. Mistakes in $colorredcanceltextcanceled red$. Corrections in $colorpurplepurple$.

$beginequation
colorblue2x = sqrtx+3colorgreenge 0\
colorblue4x^2 = x+3colorgreentext!AND! xge 0\
4x^2 - x -3 = 0 \
x^2 - frac14 x - frac34 = 0 \
x^2 - frac14x = frac34 \
colorbluex^2 -2cdotfrac 18=frac 34\
colorbluex^2 -2cdotfrac 18 +(frac 18)^2=frac 34+frac 164\
colorredcancel(x - frac12 )^2 = 1colorpurple(x - frac18)^2 = frac 4964 \
colorbluex-frac 18=pm frac 78\
colorredcancelx = frac32 , -frac12colorpurplex = 1 , -frac34colorgreentext!AND! xge 0\
colorpurplex = 1\
endequation$