# Quadratic equation and inequality, why is this method wrong?

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?

• if $x^2 \lt y^2 \implies \boxed { -y \lt x \lt y}$ – The Demonix _ Hermit Oct 19 at 13:40
• "squaring" is not a good method. From the correct $-2<-1$, squaring would yield the incorrect $4 < 1$. – GEdgar Oct 19 at 21:44
• @TheDemonix_Hermit Is it really "implies" and not equivalent $\iff$? – JDoeDoe Oct 20 at 4:41

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 $$x-3\color{red}{>}-\sqrt 5.$$

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

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

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?

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

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, $$\;A^2, $$\;$$so $$(x-3)^2<5\iff|x-3|<\sqrt 5\iff 3-\sqrt 5