Quadratics Homework Help: Solving for Specific Points in Time

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    Homework
In summary, to solve for specific points in time, you need to set the vertical displacement (y) to the desired value and then solve the resulting equation for x. This can be done using the quadratic equation or by factoring the expression. When factoring, be careful to correctly factor out the coefficient (-5) and adjust the constant term accordingly to maintain consistency with the original equation.
  • #1
benjaminxx12
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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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  • #2


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]
 
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  • #3


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
 
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  • #4


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.
 

FAQ: Quadratics Homework Help: Solving for Specific Points in Time

What are the basic concepts of quadratic equations?

Quadratic equations are algebraic equations in the form of ax^2 + bx + c = 0, where a, b, and c are constants. The highest power of the variable x is 2, and this is what makes it a quadratic equation. The graph of a quadratic equation forms a parabola, which is a U-shaped curve.

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