solve for (x+4)2 =x(x-4) +5

eddy2017

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Hi, my name is Eddy. I need help in identifying what type of equation is this one and please, help solving it step by step, if it is possible.
(x+4)2 =x(x-4) +5
thanks a lot in advance


this is the part that i don't really understand so far
The x term is -16x. Take half its coefficient (-8).
Square it (64) and add it to both sides

is this a constant.does it have to be done like that?. always?

Add '64' to each side of the equation.
-16x + 64 + x2 = 3 + 64

Reorder the terms:
64 + -16x + x2 = 3 + 64

Combine like terms: 3 + 64 = 67
64 + -16x + x2 = 67

Factor a perfect square on the left side: (don't get this part!!!) how do i factor a perfect square???
(x + -8)(x + -8) = 67


Calculate the square root of the right side: 8.185352772
 
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Hi, my name is Eddy. I need help in identifying what type of equation is this one and please, help solving it step by step, if it is possible.
(x+4)2 =x(x-4) +5
thanks a lot in advance

(x+4)2 =x(x-4) +5

x^2 + 8x + 16 = x^2 - 4x + 5

continue...

What are your thoughts?

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/announcement.php?f=33
 
(x+4)2 =x(x-4) +5

(x+4)2 =x(x-4) +5

x^2 + 8x + 16 = x^2 - 4x + 5

continue...

What are your thoughts?

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/announcement.php?f=33

Thanks so much Mr Khan:
i have found this, but i would like you to tell me if there is a shorter way to go at this.

Simplifying
(x + 4) * 2 = x(x + -14) + 5

Reorder the terms:
(4 + x) * 2 = x(x + -14) + 5

Reorder the terms for easier multiplication:
2(4 + x) = x(x + -14) + 5
(4 * 2 + x * 2) = x(x + -14) + 5
(8 + 2x) = x(x + -14) + 5

Reorder the terms:
8 + 2x = x(-14 + x) + 5
8 + 2x = (-14 * x + x * x) + 5
8 + 2x = (-14x + x2) + 5

Reorder the terms:
8 + 2x = 5 + -14x + x2

Solving
8 + 2x = 5 + -14x + x2

Solving for variable 'x'.

Reorder the terms:
8 + -5 + 2x + 14x + -1x2 = 5 + -14x + x2 + -5 + 14x + -1x2

Combine like terms: 8 + -5 = 3
3 + 2x + 14x + -1x2 = 5 + -14x + x2 + -5 + 14x + -1x2

Combine like terms: 2x + 14x = 16x
3 + 16x + -1x2 = 5 + -14x + x2 + -5 + 14x + -1x2

Reorder the terms:
3 + 16x + -1x2 = 5 + -5 + -14x + 14x + x2 + -1x2

Combine like terms: 5 + -5 = 0
3 + 16x + -1x2 = 0 + -14x + 14x + x2 + -1x2
3 + 16x + -1x2 = -14x + 14x + x2 + -1x2

Combine like terms: -14x + 14x = 0
3 + 16x + -1x2 = 0 + x2 + -1x2
3 + 16x + -1x2 = x2 + -1x2

Combine like terms: x2 + -1x2 = 0
3 + 16x + -1x2 = 0

Begin completing the square. Divide all terms by
-1 the coefficient of the squared term:

Divide each side by '-1'.
-3 + -16x + x2 = 0

Move the constant term to the right:

Add '3' to each side of the equation.
-3 + -16x + 3 + x2 = 0 + 3

Reorder the terms:
-3 + 3 + -16x + x2 = 0 + 3

Combine like terms: -3 + 3 = 0
0 + -16x + x2 = 0 + 3
-16x + x2 = 0 + 3

Combine like terms: 0 + 3 = 3
-16x + x2 = 3

The x term is -16x. Take half its coefficient (-8).
Square it (64) and add it to both sides.

Add '64' to each side of the equation.
-16x + 64 + x2 = 3 + 64

is this way correct ?. can i continue in this manner?. Thanks so much
 
All the steps shown here are correct. At a more advanced level, several steps might be combined into one, but there's no harm in carefully doing every single step one-by-one, if it helps prevent mistakes. However, I'm quite confused as to what's going on now. There seems to be two separate problems being worked on here...

In your first post, you posted (x+4)^2 =x(x-4) +5. But then in your second post, you began with the exercise (x + 4) * 2 = x(x - 14) + 5. Clearly, these two are different problems. Beginning from the problem in your second post, the steps will eventually end up at (x - 8)(x - 8) = 67, from which you can proceed to find the two solutions for x. For the problem in your original post, however, these are not the correct answers.

As for the source of your confusion in your first post: "Factor a perfect square on the left side: (don't get this part!!!) how do i factor a perfect square???" I'm not entirely sure why you're confused here, if you've followed every step up until this point. When talking about polynomial expressions, the definition of a perfect square is that it can be written exactly as the square of another polynomial. Thus, to factor a perfect square, you simply write it as (something) squared. From there, you can square root both sides.

\(\displaystyle x^2 - 16x + 64 = 67\)

\(\displaystyle (x-8)^2 = 67\)

\(\displaystyle |x-8| = \sqrt{67}\)

\(\displaystyle x - 8 = \pm\sqrt{67}\)
 
This is very confusing.

In your first post, you said the problem was to solve:

\(\displaystyle (x + 4)^2 = x(x - 4) + 5.\)

In the title of your first post, you implied the problem was to solve:

\(\displaystyle (x + 4) * 2 = x(x - 4) + 5.\)

Also, in your first post, you imply that the problem is something completely different when you say that minus 16 is the coefficient of x. Where did that come from?

Then in your second post you say you simplified the problem to:

\(\displaystyle (x + 4) * 2 = x\{x + (-\ 14)\} + 5.\)

We cannot help if we do not know what the problem is. Please write down the exact problem CAREFULLY and show us your working as far as you can go. Then we have a clue where to begin.
 
Assuming the problem is

\(\displaystyle (x + 4) * 2 = x(x - 14) + 5 \implies 2x + 8 = x^2 - 14x + 5 \implies \)

\(\displaystyle x^2 - 14x - 2x = 8 - 5 \implies x^2 - 16x = 3.\)

Getting to this point does not require nearly as many steps as you are going through.

Now what is the logic of the next step.

\(\displaystyle \left ( x + \dfrac{b}{2} \right)^2 = x^2 + bx + \left ( \dfrac{b}{2} \right )^2.\)

DO YOU SEE WHY?

\(\displaystyle \therefore x^2 + bx = x^2 + bx + 0 = x^2 + bx + \left ( \dfrac{b}{2} \right)^2 - \left ( \dfrac{b}{2} \right )^2 = \left ( x + \dfrac{b}{2} \right )^2 - \left ( \dfrac{b}{2} \right )^2.\)

DOES THAT MAKE SENSE? So

\(\displaystyle x^2 - 16x = 3 \implies \left (x + \dfrac{-\ 16}{2} \right )^2 - \left ( \dfrac{-\ 16}{2} \right )^2 = 3 \implies\)

\(\displaystyle (x - 8)^2 - (-\ 8)^2 = 3 \implies ( x - 8)^2 - 64 = 3 \implies (x- 8)^2 = 67 \implies\)

\(\displaystyle x - 8 = \pm \sqrt{67} \implies x = 8 + \sqrt{67} \text { or } x = 8 - \sqrt{67}.\)

Is there any part of this that is hazy for you? If so, please speak up.
 
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what type of equatiion

(x+4)2 =x(x-4) +5
Mr Khan and Mr jeff and the others who tried to help with this problem.
Sorry for the delay in answering your questions. I was out of the the country for a month or so.
this is the equation. I A made a mistake when I wrote the 2 not powered. I am sorry.
(x+4)2 =x(x-4) +5
if you still can break it down for me, I would be most grateful.
Thanks
 
is this way correct ? can i continue in this manner?
Is that your work? It looks like something copied and pasted from an online machine tutor (programmed by somebody who doesn't understand tutoring). Many of those steps are completely unnecessary. For example, why is a(c+b) easier to multiply than a(b+c)? Repeatedly reordering terms is silly.

I … made a mistake when I wrote the 2 [as a factor instead of an exponent] …

(x+4)2 =x(x-4) +5
We can use the caret symbol to show exponents:

(x + 4)^2

That's easier to type than formatting a superscript character.


if you still can break it down for me, I would be most grateful.
Subhotosh started it for you, by multiplying out each side. You did not answer his question.

(x+4)2 =x(x-4) +5

x^2 + 8x + 16 = x^2 - 4x + 5

What are your thoughts?

Are you taking a math class?
 
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