Why can I not take the positive or negative square root while solving a quadratic equation in this instance?

Question

I have the problem $4x^2 -36x = -81$, and I am asked to solve it by factoring. I move the $81$ to the left side of the equation, then divide by $4$, giving me $x^2 +9x + \frac {81}4 = 0$. Since I know that $\frac {81}4$ is the perfect square of $\frac 92$, I can factor this equation into $(x - \frac 92)^2 = 0$. I then take the square root, and the equation becomes $\pm$$(x-\frac 92) = 0$. I turn the equation into 2 different ones: $(x - \frac 92) = 0$ $-1(x - \frac 92) = 0$

So, that means that $x = \frac 92$.

But, $-1(x - \frac 92) = (\frac 92 - x) = 0$, so this means that x also equals $-\frac 92$. I put this through a python program I wrote, and it gave me $4.5$ or $4.5$. Where did the negative come from?

Answer 1

I would say you began to go off track when you said

I then take the square root, and the equation becomes $±(x−\frac{9}{2})=0$.

When you have $(x-\frac{9}{2})^2=0$, instead of taking the square root, you should expand the binomial:

$$(x-\frac{9}{2})(x-\frac{9}{2})=0$$

So $x=\frac{9}{2}$. This is sometimes called a double root.

By mistakenly taking the square root, you introduced the negative term, which was never there in the first place.

You might have been able to see your mistake by graphing the equation. This is a parabola that opens up with the vertex at $(\frac{9}{2},0)$. (A parabola with a vertex on the $x$-axis always has a double root.) The parabola does not cross the $x$-axis at $x=-\frac{9}{2}$. So the graph would have told you the negative answer was incorrect.

Answer 2

Notice there is an issue in your final deduction:

From $y^2 = 0$, you cannot have $y = -y$ in general (and that's what you did)

Instead, if $y^2 = 0$, you could only get $y = 0$

But of course, when you get $y=0$, $y=-y$ will hold, becase $0 = -0$

So you see the sequence of the logic?

Answer 3

Your mistake is where you say that $-(x-\frac 92)=0 \Rightarrow x=-\frac 92$.

In fact $-(x-\frac 92)=0 \Rightarrow -x+\frac 92=0 \Rightarrow x=\frac 92$

This is an example of a 'repeated root.'

Answer 4

$4x^2-36x=-81$

You added 81 to both sides:

$4x^2-36x+81 = 0$

You divided both sides by 4:

$x^2-9x+\frac{81}{4} = 0$

You factored the LHS:

$(x-\frac{9}{2})^2 = 0$

The next thing you did was incorrect. The result after taking squares roots is $x-\frac{9}{2} = ±0$, not $±(x-\frac{9}{2}) = 0$.

$x-\frac{9}{2} = ±0$

$x = \frac{9}{2} ± 0$

$x_1 = \frac{9}{2} + 0$ and $x_2 = \frac{9}{2} - 0$

$x_1 = x_2 = \frac{9}{2}$

The root is repeated.

By the way, $\frac{9}{2}$ is the solution of $\frac{9}{2} - x = 0$. Where did you get $-\frac{9}{2}$?