Question

PLEASE HELP!
COMPLETE THE SQUARE!
x^2-7x=0

Answer 1

Recall that if you have a polynomial of the form\[x^2+bx+c\]then when you complete the square you add and subtract the value \(b^2/4\). In this case, can you tell me what your value of \(b\) is?

Answer 2

-7

Answer 3

Right. So to complete the square we would have to add\[\frac{(-7)^2}{4}=\frac{49}{4}.\]This results in\[x^2-7x+\frac{49}{4}-\frac{49}{4}=0\]or equivalently\[x^2-7x+\frac{49}{4}=\frac{49}{4}.\]

Answer 4

i got that so far

Answer 5

Good.

Now you have to remember/notice that\[x^2+bx+\frac{b^2}{4}=\left(x+\frac{b}{2}\right)^2.\]Applying that to your problem, since \(b=-7\), we get\[x^2-7x+\frac{49}{4}=\left(x-\frac{7}{2}\right)^2=\frac{49}{4}.\]Still following?

Answer 6

yup!

Answer 7

Excellent. So now we can take the square root of both sides. Thus,\[x-\frac{7}{2}=\pm\sqrt{49/4}=\pm\frac72.\]This leads to two possibilities:\[x-\frac{7}{2}=\frac{7}{2}\implies x=\frac{7}{2}+\frac72=7\]or\[x-\frac72=-\frac72\implies x=0.\]So your two roots will be \(x=7\) and \(x=0\).

Answer 8

shoot i wish i couldve done that! can u help me with a few more? ur smart

Answer 9

I've got a bit of time to help out. Make sure you've tried the problems on your own first though!

Answer 10

oh i have! these are problems i was unable to solve

Answer 11

the height y of a parabolic arch is given by y=-1/16x^2+40, where x is the horizonal distance from the center of the base of the arch. All distances are in feet!

A.) what is the highest point of the arch?
B.) How wide is the arch at the base to the nearest tenth of a foot?

Answer 12

i have no idea
how to do this

Answer 13

First you need to remember what the formula is for the center of a parabola. If your equation is\[ax^2+bx+c,\]then the center is given by \(x=-b/2a\). In your case, \(b=0\), so that the center will be at \(x=0\). So we're halfway to finding the maximum height. To find the max height, we need to plug this value back into the equation. So\[y(0)=\frac{-1}{16}0^2+40=40 \;ft.\]is the maximum height. Make sense so far?

Answer 14

poo kiddo... completing the square on this problem, Lmao .

Answer 15

yes so far yes.

Answer 16

Good. To find the width at the base, you have to realize that the base is where the height is 0. So you need to find the roots of your equation. Can you do that?

Answer 17

hmm im not sure...im trying.. so the highest point is 40 but the hieght is still 0?

Answer 18

The highest point is 40, but that was part a. Now we're focusing on the second part. And kind of by definition of what an arch is, the base of the arch should be at height 0. It will meet the ground at two points, say \(r_1\) and \(r_2\). Then the distance between these points will merely be \(r_1-r_2\) (or possibly \(r_2-r_1\) if the plus/minus signs don't work out).

So to find the points where the arch meets the ground, we need to set height equal to 0.

Answer 19

ive been trying all kinds of ways to get this...ugh i suck at math

Answer 20

ok ok ..so how do we do that? put it into our formula?

Answer 21

What formula? The quadratic formula?

Answer 22

i guess ya

Answer 23

That would work just fine (although is a slightly faster way in this case). Can you tell me what you get for the roots?

Answer 24

huh? u have to show me i dont understand

Answer 25

like where would u put the 40?

Answer 26

So your equation is \[-\frac{x^2}{16}+40\]To find the roots we set it equal to 0\[-\frac{x^2}{16}+40=0\]So\[40=\frac{x^2}{16}\]Take the square root of both sides, and\[\pm\sqrt{40}=\pm2\sqrt{10}=\frac{x}{4}\]So\[x=\pm8\sqrt{10}\]Did you follow all this?

Answer 27

the answer is suppsed to be 50.6 but idk how the answer key got it

Answer 28

We're almost there. Now we find the distance between those two values. The distance is just\[8\sqrt{10}-(-8\sqrt{10})=8\sqrt{10}+8\sqrt{10}=16\sqrt{10}.\]When you use a calculator to find this value numerically, we get\[16\sqrt{10}\approx50.6\]

Answer 29

see i did this all wrong, i did it quadratic formula way and ya it didnt work

Answer 30

The quadratic formula way should still work, but there's more chance for error.

Answer 31

i think i probably did then, i like your way better.my teacher didnt teach me a good way ):

Answer 32

That's too bad... A lot of this just comes with practice, experience, and seeing other people do the same thing before me though.

Answer 33

oh really? hmm. i just need more practice. Do you know how to solve this? (8-3i)(6+9i)??? you Foil right?

Answer 34

Right. Just remember that \(i^2=-1\).

Answer 35

i got 75+54i?

Answer 36

Looks right.

Answer 37

how do you solve y=x^2-11x+24 y=x-3 Solve each system

Answer 38

Why don't you post this in a new thread now so that we can start with a blank slate again.

Answer 39

ok