Correcting the Mistakes in Completing the Square for y=3x^2+2x-1

  • Thread starter thomasrules
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In summary, the correct solution for putting y=3x^2+2x-1 into y=a(x-h)^2+K is y=3(x-1/3)^2-4/3. The mistake in the previous solution was using ah^2-1 instead of ah^2+k in the second equation.
  • #1
thomasrules
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Ok so can you please tell me where I go wrong here. I want to put
y=3x^2+2x-1 into y=a(x-h)^2+K

ax^2-2ah+ah^2+k=3x^2-2x+ah^2-1

ax^2=3x^2 -----> a=3
-2ahx=-2x
-2(3)hx=-2x ----->h=1/3
K=----->-1

Therefore:

y=3(x-1/3)^2-1

But I think it's suppost to be: y=3(x+1/3)^2-4/3

Is that right? So what is wrong...
 
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  • #2
thomasrules said:
Ok so can you please tell me where I go wrong here. I want to put
y=3x^2+2x-1 into y=a(x-h)^2+K

ax^2-2ah+ah^2+k=3x^2-2x+ah^2-1

ax^2=3x^2 -----> a=3
-2ahx=-2x
-2(3)hx=-2x ----->h=1/3
K=----->-1

Therefore:

y=3(x-1/3)^2-1

But I think it's suppost to be: y=3(x+1/3)^2-4/3

Is that right? So what is wrong...

There's nothing wrong with your method...except that I find it very confusing! :confused:

I want to put
y=3x^2+2x-1 into y=a(x-h)^2+K

ax^2-2ah+ah^2+k=3x^2-2x+ah^2-1
I know you are trying to compare terms to find h and k here, but how did the 2x get a negative sign?

I'd recommend looking at it like this:
[tex]y=3x^2+2x-1[/tex]
[tex]y=3x^2+2x+D-D-1[/tex]

Now, a perfect square in the form [tex]a(x-h)^2[/tex] looks like:
[tex]ax^2-2ahx+ah^2=3x^2+2x+D[/tex]
We see that -2ah=2 and ah^2=D.
We of course know what a is, right? I leave it to you to finish finding h and D.

Then you've got:
[tex]y=(3x^x+2x+D)-D-1[/tex]
[tex]y=a(x-h)^2-D-1[/tex]
with whatever D value you have.

-Dan
 
  • #3
y=3x^2+2x-1 into y=a(x-h)^2+K

ax^2-2ah+ah^2+k=3x^2-2x+ah^2-1

In going from the first line to the second, it looks quite screwed up.
 
  • #4
Yea thanks Dan for your help I now clearly understand it..

Btw you that was a mistake with the -2 , what should have bee +2
 
  • #5
thomasrules said:
Ok so can you please tell me where I go wrong here. I want to put
y=3x^2+2x-1 into y=a(x-h)^2+K

ax^2-2ah+ah^2+k=3x^2-2x+ah^2-1
This is where you went wrong: you want
[tex]ax^2- 2ah+ ah^2+ k= 3x^2- 3x-1[/tex]
not "ah2- 1".


ax^2=3x^2 -----> a=3
-2ahx=-2x
-2(3)hx=-2x ----->h=1/3
K=----->-1
No, ah2+ k= -1. Since a= 3 and h= 1/3, ah2= 1/3 and k= -1- 1/3= -4/3.

Therefore:

y=3(x-1/3)^2-1

But I think it's suppost to be: y=3(x+1/3)^2-4/3

Is that right? So what is wrong...
 

FAQ: Correcting the Mistakes in Completing the Square for y=3x^2+2x-1

What is completing the square?

Completing the square is a mathematical technique used to solve quadratic equations. It involves manipulating the equation to create a perfect square trinomial, which can then be easily solved.

Why is completing the square important?

Completing the square is important because it allows us to solve quadratic equations that cannot be easily solved by factoring or using the quadratic formula. It also helps us to find the maximum or minimum value of a quadratic function, which has many real-world applications.

Can you walk me through the steps of completing the square?

Sure! The first step is to move the constant term to the right side of the equation. Then, take half of the coefficient of the x term and square it. Add this value to both sides of the equation. Next, factor the perfect square trinomial on the left side. Finally, take the square root of both sides and solve for x.

When should I use completing the square instead of the quadratic formula?

You should use completing the square when the quadratic equation cannot be easily factored and the coefficients are not nice numbers. This method also allows you to find the maximum or minimum value of a quadratic function, which is not possible with the quadratic formula.

Can completing the square be used for equations with a degree higher than 2?

No, completing the square is only applicable to quadratic equations, which have a degree of 2. For equations with a degree higher than 2, other methods such as the cubic formula or numerical methods must be used.

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