Completing the square and Vertex Form

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Ghost

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Hi, I need some help. I have an upcoming exam, and I am trying to find the best method to complete the square for quadratics with an x^2 coefficient of more than one and representing it in its vertex form.

For instance, let's say the equation,
2x^2 - 5x + 2 = 0.

I learnt about factoring out the the coefficient from the x-terms and adding and subtracting a new term in the equation to get something like,
2(x + 5/4)^2 - 9/8 = 0

However, I am seeing some online tutorials that divide the entire equation by the coefficient and complete the square. The equation becomes,
x^2 - 5/2x + 1 = 0
and they end up with,
(x + 5/4)^2 - 9/16 = 0

Both methods may give the same roots but on a graph the curves will not be the same. Is the first method only for the vertex form? Is the second method only for solving? Can someone pls explain, it would be really helpful! Thanks
 
Ghost said:
Hi, I need some help. I have an upcoming exam, and I am trying to find the best method to complete the square for quadratics with an x^2 coefficient of more than one and representing it in its vertex form.

For instance, let's say the equation,
2x^2 - 5x + 2 = 0.

I learnt about factoring out the the coefficient from the x-terms and adding and subtracting a new term in the equation to get something like,
2(x + 5/4)^2 - 9/8 = 0

However, I am seeing some online tutorials that divide the entire equation by the coefficient and complete the square. The equation becomes,
x^2 - 5/2x + 1 = 0
and they end up with,
(x + 5/4)^2 - 9/16 = 0

Both methods may give the same roots but on a graph the curves will not be the same. Is the first method only for the vertex form? Is the second method only for solving? Can someone pls explain, it would be really helpful! Thanks
Click to expand...
Neither of those is correct, though close.
Show the steps.
 
I had reposted the question. Here is a better version.

When solved, both method may give the same roots, but the curves are different. Are both methods used for the vertex form or just the first?

For instance, [math]2x^2 - 5x + 2[/math]
The first method factors out the the coefficient of both x-terms and adds/subtracts a new term,

[math]2x^2 - 5x + 2[/math]
[math]2(x^2 - {5 \over 2} + 1)[/math]
[math]2(x^2 - {5 \over 2} + ({5 \over 4})^2 + 1 - ({5 \over 4})^2)[/math]
[math]2((x - {5 \over 4})^2 - {9 \over 16})[/math]
[math]2(x^2 - {5 \over 4})^2 - {9 \over 8}[/math]

The second method divides the quadratic by the coefficient, then completes the square,

[math]2x^2 - 5x + 2[/math]
[math]x^2 - {5 \over 2} + 1[/math]
[math]x^2 - {5 \over 2} + ({5 \over 4})^2 + 1- ({5 \over 4})^2 [/math]
[math](x^2 - {5 \over 4})^2 - {9 \over 16}[/math]
 
Ghost said:
I had reposted the question. Here is a better version.

When solved, both method may give the same roots, but the curves are different. Are both methods used for the vertex form or just the first?

For instance, [math]2x^2 - 5x + 2[/math]
The first method factors out the the coefficient of both x-terms and adds/subtracts a new term,

[math]2x^2 - 5x + 2[/math]
[math]2(x^2 - {5 \over 2} + 1)[/math]
[math]2(x^2 - {5 \over 2} + ({5 \over 4})^2 + 1 - ({5 \over 4})^2)[/math]
[math]2((x - {5 \over 4})^2 - {9 \over 16})[/math]


The second method divides the quadratic by the coefficient, then completes the square,

[math]2x^2 - 5x + 2[/math]
[math]x^2 - {5 \over 2} + 1[/math]
[math]x^2 - {5 \over 2} + ({5 \over 4})^2 + 1- ({5 \over 4})^2 [/math]
[math](x^2 - {5 \over 4})^2 - {9 \over 16}[/math]
Click to expand...
[math]2\left(x - {5 \over 4}\right)^2 - {9 \over 8} = 0\\ \text{Dividing by 2 yields the same results as method 2}\\ \left(x - {5 \over 4}\right)^2 - {9 \over 16} = 0\\[/math]
The difference was when the 2 gets divided, both are valid approaches to solving quadratic equations.
 
Last edited:
Ghost said:
Hi, I need some help. I have an upcoming exam, and I am trying to find the best method to complete the square for quadratics with an x^2 coefficient of more than one and representing it in its vertex form.

For instance, let's say the equation,
2x^2 - 5x + 2 = 0.

I learnt about factoring out the the coefficient from the x-terms and adding and subtracting a new term in the equation to get something like,
2(x - 5/4)^2 - 9/8 = 0

However, I am seeing some online tutorials that divide the entire equation by the coefficient and complete the square. The equation becomes,
x^2 - 5/2x + 1 = 0
and they end up with,
(x - 5/4)^2 - 9/16 = 0

Both methods may give the same roots but on a graph the curves will not be the same. Is the first method only for the vertex form? Is the second method only for solving? Can someone pls explain, it would be really helpful! Thanks
Click to expand...
Yes, the second method, which changes the function on the LHS, is convenient when you are solving an equation (as you are doing here), because all that matters is that the two sides remain equal.

The first is needed in other situations, where you are graphing a function like y = 2x^2 - 5x + 2, and dividing would mess it up. (But you could do it when you are solving, if you choose.)
 
Ghost said:
The second method divides the quadratic by the coefficient, then completes the square,

1)[math]2x^2 - 5x + 2[/math]2)[math]x^2 - {5 \over 2} + 1[/math]3)[math]x^2 - {5 \over 2} + ({5 \over 4})^2 + 1- ({5 \over 4})^2 [/math]4[math](x^2 - {5 \over 4})^2 - {9 \over 16}[/math]
Click to expand...
Although lines 2, 3, 4 and 5 are exactly the same, lines 1 and 2 are not the same!
Pick any value of your choice for x and you'll get the same result whether you use line 2, 3 or 4. However, for (almost) all values you pick for x, lines 1 and 2 will yield different results.
What can you say about the x-values which yield the same result for the expression in line 1 or for line 2?
 
Last edited:
Thanks guys, sorry about the typos, still new to the math input.

Yes, I tested on graphing software and saw that the curve changes when the quadratic is divided by the coefficient. It cuts the y-axis at a different point too. Only the roots are the same. I'll stick with the first method.
 
Last edited:
Ghost said:
Yes, I tested on graphing software and saw that the curve changes when the quadratic is divided by the coefficient. It cuts the y-axis at a different point too. Only the roots are the same. I'll stick with the first method.
Click to expand...
Keep in mind, though, that you can divide when you are solving an equation, which is the problem you were asking about. It's important to be aware of your goal, which determines what you can do; and in solving, anything that can make numbers smaller, or eliminate fractions, or in this case parentheses, can make the work easier. In general, I recommend it. Graphing has a different set of requirements to keep in mind.
 
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