Quadratics homework help

  • #1

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
 

Answers and Replies

  • #2
14
0


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:
  • #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
 
Last edited:
  • #4
14
0


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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