Quadratic Equation - Solving for x

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Homework Help Overview

The discussion revolves around solving the quadratic equation x - 3 = -x². Participants are attempting to manipulate the equation into a standard form and explore various methods for finding the roots.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss different algebraic manipulations to rearrange the equation, with some suggesting the quadratic formula and others attempting to complete the square. There are questions about the correctness of initial steps and the application of algebraic principles.

Discussion Status

The conversation is ongoing, with participants providing feedback on each other's approaches. Some guidance has been offered regarding the quadratic formula and the importance of correctly setting up the equation, but there is no explicit consensus on the best method to proceed.

Contextual Notes

One participant expresses a lack of familiarity with the quadratic formula, indicating a potential gap in foundational knowledge that may affect their ability to engage fully with the problem. There are also mentions of differing teaching methods and assumptions about prior knowledge.

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

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

and then you can use general solution to quadratic equation OR complete square.
 
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?
 
Last edited:
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

[tex]ax^{2}+bx+c=0[/tex] 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!

[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
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.

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