Quadratics Homework Help: Solving for Specific Points in Time

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Homework Help Overview

The discussion revolves around solving a quadratic equation to find specific points in time when an object is at a height of 6 meters. The equation provided is -5x^2 + 10x + 2 = 6, which participants are attempting to manipulate to find the values of x that satisfy this condition.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the method of setting the equation equal to 0 and factoring as a means to solve for x. There is also mention of using the quadratic formula. Questions arise regarding the manipulation of the equation and the introduction of constants in the factoring process.

Discussion Status

Some participants have provided guidance on how to approach the problem, including correcting a mistake in factoring and suggesting the use of the quadratic formula. There is an acknowledgment of the correctness of the revised approach, but questions about specific constants remain open for exploration.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is an emphasis on understanding the manipulation of the quadratic equation rather than simply arriving at a solution.

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Homework Statement



I am meant to look for both points when thiobject is 6m above the ground

the equation is

-5x^2+10x+2=6

my question is how do i solve it for specific points in time,
Do i just make the y term 0 and solve from there?



Homework Equations





The Attempt at a Solution



-5x^2 +10x -4

-5(x-1)^2-5=0
 
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I'm guessing [itex]y[/itex] is the vertical displacement and [itex]x[/itex] is some other parameter (the point or points you want to find at which [itex]y = 6[/itex]m). Thus you need to solve for [itex]x[/itex] given [itex]y = 6[/itex]m.

So all you need to do is solve the equation for [itex]x[/itex] like you have already started doing (either using the quadratic equation or by factorizing like you have done). You have however made a slight mistake in your factoring.

It should be

[itex]-5{(x-1)}^2 + 1=0[/itex]

Multiply this out and you should see that this gives you the original expression.

From this it is clear that (subtract 1 and divide by -5 on both sides).

[itex]{(x-1)}^2 = \frac{1}{5}[/itex]

[itex]x-1 = \pm \sqrt{\frac{1}{5}}[/itex]

[itex]x = 1 \pm \sqrt{\frac{1}{5}}[/itex]

You could also have used the quadratic equation:

[itex]x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}[/itex]

To solve the general quadratic: [itex]ax^2 + bx + c = 0[/itex].
Thus in your case we solve for x like so...

[itex]x = \frac{-10 \pm \sqrt{10^2-4(-5)(-4)}}{2(-5)}[/itex]

[itex]= \frac{-10 \pm \sqrt{100-80}}{-10}[/itex]

[itex]= \frac{-10 \pm \sqrt{20}}{-10}[/itex]

[itex]= 1 \mp \sqrt{\frac{20}{100}}[/itex]

[itex]= 1 \mp \sqrt{\frac{1}{5}}[/itex]
 
Last edited:


Thank you for the help, I am just curious as to where you get the +1 constant from.

nvm, after revision i found your answer to be correct.
You have my gratitude. :D
 
Last edited:


In case the +1 is still bothering you, consider first the expression that you ended up with.

[itex]-5{(x-1)}^2-5=0[/itex]

This expression should be consistent with the original, when one multiplies out to see if it does agree with the initial equation one gets.

[itex]-5x^2 -5(-2x) -5(1) -5= 0[/itex]

[itex]-5x^2 +10x -5 -5= 0[/itex]

We see now that in order to get -4 and not -10 we need to replace the -5 with a +1.
 

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