Quadratic Equation - Solving for x

In summary, the conversation is about solving the equation x-3=-x^2 and the correct steps to get the solution. The initial attempt was incorrect and the correct first step is to obtain x^2 + x - 3 = 0. The conversation then discusses the use of the quadratic formula and the general form of a quadratic equation. The correct solution is (x+1/2)^2 = 13/4 and the mistake in the initial attempt was not multiplying both the denominator and numerator by 4. The conversation ends with a reminder to not be lazy when solving equations.
  • #1
Larrytsai
228
0
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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  • #2
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?).
 
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  • #3
-1x2 = 3?
 
  • #4
Trying again: Your first steps were wrong. You should first obtain

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

and then you can use general solution to quadratic equation OR complete square.
 
  • #5
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?
 
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  • #6
sry I am only in gr 11 and this was how i was taught... and that formula was never taught to me
 
  • #7
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?
 
  • #8
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
 
  • #9
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!
 

FAQ: Quadratic Equation - Solving for x

What is a quadratic equation?

A quadratic equation is a polynomial equation of degree 2, which means the highest exponent of the variable is 2. It can be written in the form ax^2 + bx + c = 0, where a, b, and c are constants and x is the variable.

How do you solve a quadratic equation?

There are several methods to solve a quadratic equation, including factoring, completing the square, and using the quadratic formula. The method you use depends on the form of the equation and your personal preference.

What is the quadratic formula?

The quadratic formula is a commonly used formula to find the solutions of a quadratic equation. It is written as x = (-b ± √(b^2 - 4ac)) / 2a, where a, b, and c are the constants in the equation ax^2 + bx + c = 0.

How many solutions can a quadratic equation have?

A quadratic equation can have either two real solutions, one real solution, or no real solutions. The number of solutions depends on the discriminant, which is b^2 - 4ac in the quadratic formula. If the discriminant is greater than 0, there are two real solutions. If the discriminant is equal to 0, there is one real solution. And if the discriminant is less than 0, there are no real solutions.

What is the difference between real and imaginary solutions?

Real solutions are solutions that are numbers on the number line, while imaginary solutions are solutions that involve the imaginary number i (√-1). In the context of quadratic equations, real solutions are the values of x that make the equation true, while imaginary solutions do not make the equation true.

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