Quadratics Homework Help: Solving for Specific Points in Time

[benjaminxx12]

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

 

[Isak BM]

I'm guessing y is the vertical displacement and x is some other parameter (the point or points you want to find at which y = 6m). Thus you need to solve for x given y = 6m.

So all you need to do is solve the equation for x 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

-5{(x-1)}^2 + 1=0

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).

{(x-1)}^2 = \frac{1}{5}

x-1 = \pm \sqrt{\frac{1}{5}}

x = 1 \pm \sqrt{\frac{1}{5}}

You could also have used the quadratic equation:

x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}

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

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

= \frac{-10 \pm \sqrt{100-80}}{-10}

= \frac{-10 \pm \sqrt{20}}{-10}

= 1 \mp \sqrt{\frac{20}{100}}

= 1 \mp \sqrt{\frac{1}{5}}

 

[benjaminxx12]

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

 

[Isak BM]

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

-5{(x-1)}^2-5=0

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

-5x^2 -5(-2x) -5(1) -5= 0

-5x^2 +10x -5 -5= 0

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