Can I Square Both Sides of an Equation
Consider the graph of $y = x^2$:
I sometimes think of a graph (or function in general) every bit: "you give me $x$, I give you $y$" -- Give me iii, I give you ix. Give me -5 I give you lot 25, and so on.
Solving $10^2 = 4$ tries to contrary this -- I give you $y$, and you take to come with the $10$ that would have produced this.
Graphically, the starting time procedure ($x \rightarrow y$) is taking vertical lines and finding the height of the graph where they intersect the red curve. The second process ($y \rightarrow x$) is the reverse -- taking horizontal lines and finding the $ten$ value (displacement from the vertical line at $x = 0$) where they intersect the red curve.
For the function $x^two$, nosotros can encounter that nearly horizontal lines will intersect the bend twice -- this leads to ambiguity about which $10$ you lot should "give" me to answer my question. To put this another way, the process of squaring a number ways that our tit-for-tat exchange (you give me $x$, I give you $y$; I give you $y$, you give me $x$) tin be broken.
Turning to the specific relationship you lot mentioned, $x = ten+1$ vs. $x^two = (ten+1)^ii$, recollect of what the latter means in terms of the graph.
To solve this nosotros would notice an $x$ so that if nosotros motion the right by one, the height of intersection is the aforementioned at both places. We tin run into how this might happen due to the "refraction" of the curve at the origin:
The equation "gains" a solution (when going from $10 = 10+1$ to $ten^2 = (x+1)^2$ considering of what happens when nosotros pass the origin and the curve bends dorsum upward (technical term: non-monotonicity).
The plot shows what happens to the $y$ value equally we jump to the right by 1 starting from -2.6 (landing at -one.6, -.6, .four, 1.iv, ii.4, ...).
Hopefully you can come across that if we had chosen a unlike value (due east.g. -2.3, -ii.011, -two.78), the exact $y$ values would be different but the pattern would be the same.
The last thing to convince yourself is that at that place'southward a "sweet spot" of where we could start jumping (between -three and -2) so that the $y$ values before and subsequently crossing $x = 0$ are exactly the same, instead of just ~kind of close~. (A bonus is to convince yourself that in that location'southward but one such sweet spot)
Hither's the R code I used to generate these plots:
Plot 1:
curve(ten^ii, -3, 3, asp = one, lwd = 3L, las = 1L, ylab = expression(x^2), col = 'reddish', main = 'Not-invertibility of Squaring') abline(five = 0, h = 0) abline(h = four, col = 'bluish', lty = 2L) axis(side = 1L, at = c(-2, two)) segments(c(-2, 2), c(0, 0), c(-2, 2), c(4, four), col = 'darkgreen', lty = 2L) Plot ii:
curve(x^2, -two, two, asp = 1, lwd = 3L, ylab = expression(x^ii), col = 'red', main = 'Inchworming the Graph', xaxt = 'n', yaxt = 'n') abline(five = 0, h = 0) x = c(-2.6, -1.6, -.half-dozen, .four, 1.4, ii.4) y = 10^2 n = length(ten) arrows(x[-n], y[-n], x[-1], y[-1]) axis(side = one, at = x) axis(side = 2, at = y, las = 1L) segments(x, rep(0, n), x, y, col = 'darkgreen', lty = 2L) abline(h = y, col = 'blue', lty = 2L) Source: https://math.stackexchange.com/questions/2821764/how-does-squaring-both-sides-of-an-equation-lead-to-extraneous-solutions
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