Find Value of Squareroot of 3: Using the Graph & Quadratic Equation

In summary: I think the substitution might not have been clear. What I mean is that maybe there was a mistake in replacing all x terms by $\sqrt{3}$.
  • #1
mathlearn
331
0
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 :)
 
Last edited:
Mathematics news on Phys.org
  • #2
y = -2x
 
  • #3
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 :)
 
  • #4
Where do the parabola and line intersect?
 
  • #5
MarkFL said:
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)
 
  • #6
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 \(\displaystyle x^2- 2x- 3= ax+ b\) so that \(\displaystyle x^2- (2+ a)x- (3+ b)= 0\). If [tex]\sqrt{3}[/tex] is a root then, in order that the coefficients be integers, [tex]-\sqrt{3}[/tex] must also be so that we must have [tex](\sqrt{3})^3- (2+ b)\sqrt{3}- (3+ b)= 3- (2+ a)\sqrt{3}- 3- b= -(2+a)\sqrt{3}- b= 0[/tex] and [tex](-\sqrt{3})^2- (2+ b)(-\sqrt{3})- 3- b= 3+ (2+ b)\sqrt{3}- 3- b= (2+a)\sqrt{3}- b= 0[/tex]. Adding those two equations, the "a" terms cancel giving -b= 0 so b= 0. Then we have [tex]-(2+ a)\sqrt{3}= 0[/tex] so that 2+ a= 0 and a= -2.
 
  • #7
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!
 
  • #8
greg1313 said:
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 :)
 
Last edited:

FAQ: Find Value of Squareroot of 3: Using the Graph & Quadratic Equation

How do you find the value of square root of 3 using a graph?

To find the value of square root of 3 using a graph, you can plot the graph of the quadratic equation y = x^2 - 3 and look for the point where y = 0. This point will give you the value of x, which is the square root of 3.

What is the significance of using the quadratic equation to find the value of square root of 3?

The quadratic equation is a mathematical formula that represents a parabola, which is the shape of the graph of y = x^2 - 3. By using this equation, we can easily calculate the value of square root of 3 without having to plot a graph.

Can you explain the steps involved in using the quadratic equation to find the value of square root of 3?

First, we need to write the quadratic equation for the given problem, which is y = x^2 - 3. Then, we can set y = 0 and solve for x. This will give us the two solutions of the equation, one of which is the value of square root of 3.

Is it possible to find the value of square root of 3 without using a graph or the quadratic equation?

Yes, it is possible to find the value of square root of 3 by using other methods such as the long division method or the Newton-Raphson method. However, the graph and quadratic equation provide a more efficient and accurate way to calculate the value.

What are the applications of finding the value of square root of 3 using the graph and quadratic equation?

The value of square root of 3 is used in various fields such as engineering, physics, and finance. It helps in solving problems related to finding areas and volumes, calculating velocities and accelerations, and estimating financial risks and returns.

Similar threads

Back
Top