Where do the strange roots of an equation come from?
Like this case:
$$x - 5 = 2$$ $$(x-5)(x+1) = 2(x+1)$$ $$x^2-4x-5=2x+2$$ $$x^2-6x-7=0$$ $$x^2-7x+x-7=0$$ $$x^2+x-7x-7=0$$ $$x(x+1)-7(x+1)=0$$ $$(x-7)(x+1)=0$$ $x = 7$ or $x = -1$
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2Here: $$(x-5)(x+1) = 2(x+1)$$ – Shaun Aug 08 '21 at 17:14
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3You cannot multiply by $,x+1,$ without first establishing that $,x+1\ne0 \iff x \ne -1,$. – dxiv Aug 08 '21 at 17:17
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@dxiv So it just came about because of that fact? – Artures Aug 08 '21 at 17:19
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1@Artures Yes. If you multiply the two sides of an equation by $0$ you always get equality $0=0$. If you multiply the two sides of the equation by $f(x)$ any root of $f(x)=0$ becomes a root of the new equation obtained after multiplication. In your case, multiplying by $x+1$ introduces the extraneous root $x=-1$. – dxiv Aug 08 '21 at 17:21
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Cf. extraneous solution – J. W. Tanner Aug 08 '21 at 18:23
2 Answers
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You introduced the "strange" root!
$$(x-5-2) = 0$$ $$(x-7) = 0$$ $$(x-7)(x+1) = 0$$ $$\textit{etc.}$$
mjw
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Multiplying the equation with $(x+1)$ on both sides was really absurd.
If you multiply an equation with a variable term, you may gain some roots and may lose some of them by dividing.
Since $0$ multiply by any number is $0$, you are changing the given equation with another one. You are just creating a different equation from existing one .
For example:- Let's say $ x^2 = x$
If we divide both sides with $x$ we will get $x=1$
Here, we missed a solution which was $0$.
And since $x^2=x$ doesn't implies that $x$ is only equal to $1$, our step of division was wrong!
And if we multiply both sides of the equation $x=1$ with $x$, we will get $x^2=x$, here we have created another solution $0$. Both equations are different from each other and our step of multiplication was wrong.
You can also visit this answer for more details.