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)
$x-1=\pm\sqrt{3}$ , \(\displaystyle \therefore\) $2.7-1=+\sqrt{3}$
$1.7=+\sqrt{3}$
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 $\sqrt{3}$.
The square root of 3 is an irrational number, meaning it cannot be expressed as a simple fraction. It is approximately 1.7320508075688772.
The square root of 3 can be found by using the roots of the graph. This involves finding the point on the graph where the line crosses the x-axis at the value of 3, and then finding the corresponding y-value. This y-value will be the square root of 3.
Yes, the square root of 3 can be approximated by using the roots of the graph method described above. This will give you a close approximation of the actual value.
The square root of 3 is an irrational number, meaning it cannot be expressed as a simple fraction. It is a non-repeating, non-terminating decimal.
The square root of 3 is a fundamental mathematical concept that is used in many calculations and formulas in various fields, such as physics, engineering, and finance. Being able to find and approximate it accurately is essential for solving complex problems and making accurate calculations.