Find and approximate value square root of 3 using the roots of the graph.

[mathlearn]

Using a graph of function $y=3-(x-1)^2$ which has got its negative & positive root s-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]

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 :)

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 $\sqrt{3}$?

 

[mathlearn]

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 $\sqrt{3}$?

If x=-0.8, I guess It should be -0.7

$x- 1=+2.7$
$x- 1+1=2.7+1$

$x- 1=+2.7$
$x=3.7$(Oops! Looks like something is wrong)(Doh) $x- 1=-0.7$
$x- 1+1=-0.7+1$

$x=0.3$(Oops! Looks like something is wrong Again !)(Doh)

I would like to know the different way's of approximating a square root using the roots of a graph. Like one I already know is using the quadratic formula

 

[Evgeny.Makarov]

$x- 1=+2.7$

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]

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 $x\approx-0.8$.

Oh! my apologies (Smile).

Then,

$x-1=\pm\sqrt{3}$
$2.7-1=\pm\sqrt{3}$
$1.7=\pm\sqrt{3}$

$0.7-1=\pm\sqrt{3}$
$-1.7=\pm\sqrt{3}$

Thanks (Smile), 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]

$2.7-1=\pm\sqrt{3}$

This is incorrect because the left-hand side is positive.

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.

I don't know of another way to solve this particular problem.

 

[mathlearn]

This is incorrect because the left-hand side is positive.

(Smile) Why is that ?, +1.7 is an approximate value for the square root of 3. (Smile) Please comment.

 

[Evgeny.Makarov]

Why is that ?

because the left-hand side is positive.

which cannot be said about $\pm\sqrt{3}$.

 

[mathlearn]

which cannot be said about $\pm\sqrt{3}$.

Hopefully, I guess this is correct now (Happy)

$x-1=\pm\sqrt{3}$ , $$\therefore$$ $2.7-1=+\sqrt{3}$
$1.7=+\sqrt{3}$

$0.7-1=- \sqrt{3}$
$-1.7=- \sqrt{3}$

 

[Evgeny.Makarov]

$0.7-1=- \sqrt{3}$
$-1.7=- \sqrt{3}$

Where did you get $0.7$ from? And $0.7-1\ne-1.7$.

More precisely, $x-1=\sqrt{3}$ when $x\approx2.7$ and $x-1=-\sqrt{3}$ when $x\approx-0.8$.

 

[HallsofIvy]

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 2.7

$0.7-1=- \sqrt{3}$
$-1.7=- \sqrt{3}$

Better: $x- 1= -\sqrt{3}$ so $x= 1- \sqrt{3}$ which is approximately -0.7.

 

[Evgeny.Makarov]

No, in the original problem $x$ is known and one is asked to estimate $\sqrt{3}$.

 

[mathlearn]

Where did you get $0.7$ from? And $0.7-1\ne-1.7$.

(Doh) Sorry , I have mistyped -0.7 as +0.7

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.

(Nod) Yes Isolating 'X' on the RHS would look neater and tidier.

No, in the original problem $x$ is known and one is asked to estimate $\sqrt{3}$.

Yes (Nod)
Many Thanks (Happy)