The discussion centers on approximating the square root of 3 using the function graph of \(y=3-(x-1)^2\). Participants identify the roots of the function as approximately -0.8 and 2.7, leading to the conclusion that \(x-1=\pm\sqrt{3}\). The correct approximations for \(\sqrt{3}\) are derived from the positive root, yielding \(x=1+\sqrt{3}\approx2.7\) and the negative root yielding \(x=1-\sqrt{3}\approx-0.7\). The quadratic formula is mentioned but deemed unnecessary for this specific problem.
PREREQUISITESMathematics students, educators, and anyone interested in understanding the graphical interpretation of quadratic equations and their roots for approximating square roots.
This problem is much easier than you are making it.mathlearn said:Using a graph of function $y=3-(x-1)^2$ which has got its negative & positive roots -0.8 and 2.7 respectively, Find an approximate value for $\sqrt{3}$.
Any suggestions on how to begin? Should I be using the quadratic formula here?
Many Thanks :)
HallsofIvy said:This problem is much easier than you are making it.
If $y= 3- (x- 1)^2= 0$ then $(x- 1)^2= 3$ so $x- 1=\pm\sqrt{3}$. If x= -0.8, what is $\sqrt{3}$? If x= 2.7, what is $\sqrt{3}$?
Where are you getting this from? You were told told that $x-1=\pm\sqrt{3}$ and not $x-1=2.7$. More precisely, $x-1=\sqrt{3}$ when $x\approx2.7$ and $x-1=-\sqrt{3}$ when $x\approx-0.8$.mathlearn said:$x- 1=+2.7$
Evgeny.Makarov said:Where are you getting this from? You were told told that $x-1=\pm\sqrt{3}$ and not $x-1=2.7$. More precisely, $x-1=\sqrt{3}$ when $x\approx2.7$ and $x-1=-\sqrt{3}$ when $x\approx-0.8$.
This is incorrect because the left-hand side is positive.mathlearn said:$2.7-1=\pm\sqrt{3}$
I don't know of another way to solve this particular problem.mathlearn said:I would like to know the different ways of doing this kind of a problem 'to find the square root using the roots of a graph'. Like one way would be to use the quadratic formula.
Evgeny.Makarov said:This is incorrect because the left-hand side is positive.
mathlearn said:Why is that ?
which cannot be said about $\pm\sqrt{3}$.Evgeny.Makarov said:because the left-hand side is positive.
Evgeny.Makarov said:which cannot be said about $\pm\sqrt{3}$.
Where did you get $0.7$ from? And $0.7-1\ne-1.7$.Evgeny.Makarov said:$0.7-1=- \sqrt{3}$
$-1.7=- \sqrt{3}$
Evgeny.Makarov said:More precisely, $x-1=\sqrt{3}$ when $x\approx2.7$ and $x-1=-\sqrt{3}$ when $x\approx-0.8$.
Better:- $x- 1= \sqrt{3}$ so $x= 1+ \sqrt{3}$ which is approximately 2.7mathlearn said:Hopefully, I guess this is correct now (Happy)
$x-1=\pm\sqrt{3}$ , $$\therefore$$ $2.7-1=+\sqrt{3}$
$1.7=+\sqrt{3}$
Better: $x- 1= -\sqrt{3}$ so $x= 1- \sqrt{3}$ which is approximately -0.7.$0.7-1=- \sqrt{3}$
$-1.7=- \sqrt{3}$
Evgeny.Makarov said:Where did you get $0.7$ from? And $0.7-1\ne-1.7$.
HallsofIvy said:Better:- $x- 1= \sqrt{3}$ so $x= 1+ \sqrt{3}$ which is approximately 2.7Better: $x- 1= -\sqrt{3}$ so $x= 1- \sqrt{3}$ which is approximately -0.7.
Evgeny.Makarov said:No, in the original problem $x$ is known and one is asked to estimate $\sqrt{3}$.