How is the equation $x_1+5x_2-\sqrt{(2x_3)} = 1$ a linear equation? The answer given in the book is, "The Equation is linear".
How can an equation involving a square root like the above equation be a linear equation?
here is the cutting of the book,
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2It isn't. But if you had $\sqrt 2 x_3$ instead of $\sqrt{2x_3}$, it would be... I suspect there was a typo. – David Mitra Dec 26 '12 at 17:16
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The original post clearly did not have $2x_3$ nested in parentheses. – David Mitra Dec 26 '12 at 17:20
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no this is not misprint you can see the cutting from the book, this is (a) part and at the bottom you can see the answer – Zia ur Rahman Dec 26 '12 at 17:22
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@DavidMitra I only put parenthesis because LaTeX cut off the top of the square root sign, which sort of defeated the purpose of the question. – Sidd Singal Dec 26 '12 at 17:23
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2It's a typesetting error, clearly. The variables were not meant to be under the radicand (if the answer is to be correct). – David Mitra Dec 26 '12 at 17:25
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@mathguy Ah. If I may suggest a TeX tip: If you want a bit of extra space, use "," or "\thinspace" at the end (these are plain TeX commands). "\sqrt{2x_3}" gives $\sqrt{2x_3}$, while "\sqrt{2x_3,}" gives $\sqrt{2x_3,}$. – David Mitra Dec 26 '12 at 17:36
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@DavidMitra Ohh...I'll definitely keep that in mind, thanks! – Sidd Singal Dec 26 '12 at 18:13
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Let's hope those are ugly, big, annoying and confusing typesetting errors ( in (a) and (f) ) and not mathematical ones by that book's author... – DonAntonio Dec 26 '12 at 18:34
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Shouldn't they have written it as $x_3\sqrt2$ to avoid this mess in the first place? – Mike Dec 26 '12 at 18:44
3 Answers
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Here is the real exercise found on Amazon...
GEdgar
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1This picture is from the latest edition. It seems in some earlier edition the erroneous statements appear (see the question itself). Nowadays textbooks have on-line errata sites. In the olden days it was not so easy... – GEdgar Dec 26 '12 at 22:32
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$$(x+5y-1)=\sqrt{2z}$$ so $$(x+5y-1)^2=2z$$ and this is not a linear equation because the order of variabes are 2.
Mikasa
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Answer to title question: It's NOT!
Your question is legitimate:
$$x_1+5x_2-\sqrt{2x_3\;} = 1\tag{1}$$
$(1)$ is not a linear equation as you suggest.
Nor is $(f)$ linear, as typeset in the image.
I suspect there was a misprint in the problem set (book), or a careless typo that the author (and/or editor) over-looked, and which was intended to be:
$$x_1 + 5x_2 - \sqrt{2}\;\cdot x_3 = 1\tag{2}$$
NOW, $(2)$ is a linear equation.
amWhy
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this is not misprint you can see the cutting of the book in the question above – Zia ur Rahman Dec 26 '12 at 17:23
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this is from elementary linear algebra by Howard anton and Chris rorres – Zia ur Rahman Dec 26 '12 at 17:24
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@ZiaurRahman: May I ask you to write for me the exact definition which book did about linearity of an equation? Thanks. – Mikasa Dec 26 '12 at 17:38
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in linear equation all variables occurs only to the first power, linear equation does not involve any product or roots, and the variables must not appear as arguments to logrithmic, trignometric or exponential functions. – Zia ur Rahman Dec 26 '12 at 17:47
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1i changed my opinion because i saw the same question in the 9th edition of this book where the square root is only on 2 – Zia ur Rahman Dec 26 '12 at 17:48
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No problem, Zia ur. I would have been perplexed too, had I encountered that question as written in the older edition! – amWhy Dec 26 '12 at 17:50