- #1

- 228

- 0

Question: x-3=-x^2

atempt: (x^2)+x = 3

x^2 + x + 1/4= 3(4) + 1/4

(x+1/2)^2= 3

root everthing

x+1/2= root 3

x=+,- root 3 -1/2

- Thread starter Larrytsai
- Start date

- #1

- 228

- 0

Question: x-3=-x^2

atempt: (x^2)+x = 3

x^2 + x + 1/4= 3(4) + 1/4

(x+1/2)^2= 3

root everthing

x+1/2= root 3

x=+,- root 3 -1/2

- #2

symbolipoint

Homework Helper

Education Advisor

Gold Member

- 5,999

- 1,094

Your initial algebraic steps were wrong. From the start, you should obtain:

EDIT: bad information was HERE.

from which the solutions are very plain (what?).

EDIT: bad information was HERE.

from which the solutions are very plain (what?).

Last edited:

- #3

- 228

- 0

-1x2 = 3???

- #4

symbolipoint

Homework Helper

Education Advisor

Gold Member

- 5,999

- 1,094

[tex] \[

x^2 + x - 3 = 0

\]

[/tex]

and then you can use general solution to quadratic equation OR complete square.

- #5

- 1,631

- 4

Well, i don't really understand what u tried to do.

[tex] x-3=-x^{2}=> x^{2}+x-3=0[/tex] Now do you could either try to factor this out, or apply directly the quadratic formula, do you know it?

[tex]x_1_,_2=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}[/tex]

In this particular case, i don't think it will factor nicely, so it is better to use the quadratic formula i provided u with.

Can u go from here?

[tex] x-3=-x^{2}=> x^{2}+x-3=0[/tex] Now do you could either try to factor this out, or apply directly the quadratic formula, do you know it?

[tex]x_1_,_2=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}[/tex]

In this particular case, i don't think it will factor nicely, so it is better to use the quadratic formula i provided u with.

Can u go from here?

Last edited:

- #6

- 228

- 0

sry im only in gr 11 and this was how i was taught... and that formula was never taught to me

- #7

- 1,631

- 4

Well, you better learn it then, because not always will you be able to factor a quadratic eq. nicely.sry im only in gr 11 and this was how i was taught... and that formula was never taught to me

The general form of a quadratic equation is

[tex] ax^{2}+bx+c=0[/tex] so now can you figure out what a,b and c are in your problem?

- #8

- 228

- 0

- #9

- 1,631

- 4

Aha i gotch ya!

[tex]x^2 + x = 3=>x^{2}+2x\frac{1}{2}+\frac{1}{4}-\frac{1}{4}=3=>(x+\frac{1}{2})^{2}=3+\frac{1}{4}=>(x+\frac{1}{2})^{2}=\frac{13}{4}[/tex]

You forgot to add that 1/4 to the 3 on your right hand side.

Well, now you know what to do right?

- #10

Integral

Staff Emeritus

Science Advisor

Gold Member

- 7,200

- 56

Where did the bolded factor of 4 come from?I know this is a simple question but for some reason im getting stumped:

Question: x-3=-x^2

atempt: (x^2)+x = 3

x^2 + x + 1/4= 3(4) + 1/4

(x+1/2)^2= 3

root everthing

x+1/2= root 3

x=+,- root 3 -1/2

You should have.

[tex] (x+ \frac 1 2)^2 = 3 \frac 1 4 [/tex]

- #11

- 228

- 0

that was just multiplying the denominator by 4 so i could add fractions

- #12

- 1,631

- 4

Well you should have multiplied then both the denominator and the numerator by 4.that was just multiplying the denominator by 4 so i could add fractions

- #13

- 228

- 0

yea i know i just was too lazy to type it out >.<

- #14

- 1,631

- 4

Well, that's why you got the wrong result then. Better not be lazy!yea i know i just was too lazy to type it out >.<

- Last Post

- Replies
- 13

- Views
- 2K

- Replies
- 6

- Views
- 743

- Replies
- 4

- Views
- 1K

- Last Post

- Replies
- 15

- Views
- 2K

- Last Post

- Replies
- 6

- Views
- 791

- Last Post

- Replies
- 9

- Views
- 3K

- Last Post

- Replies
- 3

- Views
- 2K

- Last Post

- Replies
- 6

- Views
- 1K

- Last Post

- Replies
- 4

- Views
- 947

- Last Post

- Replies
- 5

- Views
- 2K