Quadratic equation and inequality, why is this method wrong?

Question

I have this inequality

$$ (x-3)^2<5 \tag1 $$

But why is the following method wrong?

Add $-5$ to both sides and expand the square \begin{align} x^2-6x+9-5&<0 \\ x^2-6x+4&<0 \end{align} And completing the square of $x^2-6x+4<0$ gives: \begin{align} x^2-6x+ (6/2)^2-(6/2)^2+4&<0 \\ (x-3)^2-3^2+4&<0 \\ (x-3)^2&<5 \\ \end{align} Squaring and adding $3$ to both sides: \begin{align} x<3\pm\sqrt 5 \tag 2 \end{align} So I have: \begin{align} x&<3+\sqrt 5 \tag 3\\ x&<3-\sqrt 5 \tag 4 \end{align} Why is this method wrong?

From the book: For $(x-3)^2<5$ we have two solutions \begin{align} x-3&<\sqrt 5 \tag 5\\ -(x-3)&<\sqrt 5 \tag 6 \end{align} So \begin{align} x&<3+\sqrt 5 \tag 7\\ x&>3-\sqrt 5 \tag 8 \end{align}

$(7)$ is the same as my $(3)$, but $(4)$ is not $(8)$. What is wrong with my method?

Answer 1

Your first few sequence of operations is completely unnecessary. Note that you eventually end up with the original inequality.

Well, so what's wrong? The problem is when you go from $(x-3)^2<5$ to $x<3\pm\sqrt 5$ by "squaring" and adding three to both sides. First, I think you meant to extract square roots, not to square. But if you do that you get this $\sqrt{(x-3)^2}<\sqrt 5,$ or in other words $$|x-3|<\sqrt 5.$$ This may be written as $$\pm(x-3)<\sqrt 5.$$ These are actually two inequalities, $x-3<\sqrt 5,$ or $-(x-3)<\sqrt 5.$ So your error is apparent in that you changed the last inequality to $x-3<-\sqrt 5.$ But that's wrong, for you should instead have $$x-3\color{red}{>}-\sqrt 5.$$

Answer 2

Well you see there is a huge difference between inequality and equality. In inequality a small change can alter sign of the inequality like multiplying by -1. Likewise here take $y^2<5$

Now if you solve like a quadratic equality it becomes $y<\sqrt{5}$ and $y<- \sqrt{5}$ Let's assume this is the solution let's take $y=-\sqrt{10}$ .This is true if we take second inequality we assumed to be true. Substitute y into it's original inequality. You'll get $10<5$ which is obviously not true . It is true if $y>- \sqrt{5}$

So for $y^2<a$ where $a>0$ inequality becomes $- \sqrt{a}<y<\sqrt{a}$

Answer 3

looking at the line before your (2) the equation can be compared to:$$ (1)^2\lt2 $$so$$ 1\lt\sqrt2$$ but $$1\not\lt-\sqrt2$$

Answer 4

From $(x-3)^2 \lt 5$ you could take the square root of both sides of the equation, giving

$$\left| x-3 \right| \lt \sqrt{5}$$ which is equivalent to $$-\sqrt{5}\lt x-3 \lt \sqrt{5}$$ $$3-\sqrt{5}\lt x \lt 3+\sqrt{5}$$

Another way of understanding what you did wrong it to ask yourself how $x$ can be less than $3-\sqrt{5}$ if $3-\sqrt{5}$ is $x$'s lower bound?

Answer 5

The following

$$(x-3)^2<5 \implies x-3<\pm\sqrt 5$$

is wrong.

What is true for $(x-3)\ge 0$ is that

$$(x-3)^2<5 \implies x-3<\sqrt 5$$

The book is using that for $a\ge 0$

$$x^2<a \implies -\sqrt a<x<\sqrt a$$

Answer 6

Your method is wrong when you translate $(x-3)^2<5$ into $x<3\pm\sqrt 5$, which means nothing.

Actually the very short way to solve the inequation uses the basic fact that, for nonnegative numbers, $\;A^2<B\iff A<\sqrt B$, $\;$so $$(x-3)^2<5\iff|x-3|<\sqrt 5\iff 3-\sqrt 5<x<3+\sqrt 5.$$