Finding √3 using a graph of x² - 2x - 3 and a straight line

[mathlearn]

There is a graph in the form of $x^2-2x-3$ determine the value of $\sqrt{3}$ to the nearest decimal place by drawing an a suitable straight line

What must be that straight line ? Usually these kind of problems are solved using the quadratic equation

Many thanks :)

 

[Greg]

y = -2x

 

[mathlearn]

Here is the desmos graph including both the parabola and the line

[graph]z2xkzb3xja[/graph]

Looking at the line $y=-2x$ I cannot exactly see what has it got to do with determining $\sqrt{3}$ to the nearest whole number

Many Thanks :)

 

[MarkFL]

Where do the parabola and line intersect?

 

[mathlearn]

Where do the parabola and line intersect?

Wow (Clapping) they intersect at -1.7 and 1.7.

What was the method used to determine the line?

Many Thanks (Smile)

 

[HallsofIvy]

I can't speak for greg1313 and MarkFL but what I would do is note that, if we take equation y= ax+ b, that line and the given quadratic will intersect where [math]x^2- 2x- 3= ax+ b[/math] so that [math]x^2- (2+ a)x- (3+ b)= 0[/math]. If $$\sqrt{3}$$ is a root then, in order that the coefficients be integers, $$-\sqrt{3}$$ must also be so that we must have $$(\sqrt{3})^3- (2+ b)\sqrt{3}- (3+ b)= 3- (2+ a)\sqrt{3}- 3- b= -(2+a)\sqrt{3}- b= 0$$ and $$(-\sqrt{3})^2- (2+ b)(-\sqrt{3})- 3- b= 3+ (2+ b)\sqrt{3}- 3- b= (2+a)\sqrt{3}- b= 0$$. Adding those two equations, the "a" terms cancel giving -b= 0 so b= 0. Then we have $$-(2+ a)\sqrt{3}= 0$$ so that 2+ a= 0 and a= -2.

 

[Greg]

Mathematics is a science and experimentation is a valuable tool. The first thing I did was to substitute $\sqrt3$ for $x$ in the given quadratic and observe the results. Get your hands dirty!

 

[mathlearn]

Mathematics is a science and experimentation is a valuable tool. The first thing I did was to substitute $\sqrt3$ for $x$ in the given quadratic and observe the results. Get your hands dirty!

Yes agreed :) By substitution I guess what was implied was replacing all x terms by $\sqrt{3}$

$x^2-2x-3$

$\sqrt{3}^2-2\sqrt{3}-3$

$3-2\sqrt{3}-3$

$-2\sqrt{3}=0=y$

And what possibly went wrong?

Many thanks :)