Quadratic Equation - Solving for x

AI Thread Summary
The discussion revolves around solving the quadratic equation x - 3 = -x^2. Participants highlight that the initial algebraic steps were incorrect and emphasize the need to rearrange the equation to x^2 + x - 3 = 0. They suggest using the quadratic formula for solutions, as factoring may not be straightforward. A misunderstanding about adding fractions is also addressed, with advice to properly multiply both the numerator and denominator to avoid errors. The conversation underscores the importance of mastering the quadratic formula for solving such equations effectively.
Larrytsai
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I know this is a simple question but for some reason I am 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
 
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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?).
 
Last edited:
-1x2 = 3?
 
Trying again: Your first steps were wrong. You should first obtain

\[<br /> x^2 + x - 3 = 0<br /> \]<br />

and then you can use general solution to quadratic equation OR complete square.
 
Well, i don't really understand what u tried to do.

x-3=-x^{2}=&gt; x^{2}+x-3=0 Now do you could either try to factor this out, or apply directly the quadratic formula, do you know it?

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

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?
 
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sry I am only in gr 11 and this was how i was taught... and that formula was never taught to me
 
Larrytsai said:
sry I am only in gr 11 and this was how i was taught... and that formula was never taught to me

Well, you better learn it then, because not always will you be able to factor a quadratic eq. nicely.

The general form of a quadratic equation is

ax^{2}+bx+c=0 so now can you figure out what a,b and c are in your problem?
 
oo i do know that that's the form i write my solution as, but for my question i stated above i don't know what I am doing wrong to keep me from reaching that formula
 
Larrytsai said:
oo i do know that that's the form i write my solution as, but for my question i stated above i don't know what I am doing wrong to keep me from reaching that formula

Aha i gotch ya!

x^2 + x = 3=&gt;x^{2}+2x\frac{1}{2}+\frac{1}{4}-\frac{1}{4}=3=&gt;(x+\frac{1}{2})^{2}=3+\frac{1}{4}=&gt;(x+\frac{1}{2})^{2}=\frac{13}{4}

You forgot to add that 1/4 to the 3 on your right hand side.
Well, now you know what to do right?
 
  • #10
Larrytsai said:
I know this is a simple question but for some reason I am 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
Where did the bolded factor of 4 come from?

You should have.

(x+ \frac 1 2)^2 = 3 \frac 1 4
 
  • #11
that was just multiplying the denominator by 4 so i could add fractions
 
  • #12
Larrytsai said:
that was just multiplying the denominator by 4 so i could add fractions

Well you should have multiplied then both the denominator and the numerator by 4.
 
  • #13
yea i know i just was too lazy to type it out >.<
 
  • #14
Larrytsai said:
yea i know i just was too lazy to type it out >.<

Well, that's why you got the wrong result then. Better not be lazy!
 
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