Dimensions of Rectangle: Solution/ Explanation

[Guest]

Here is a variation leading to a quadratic:

A rectangular piece of land has the length exceeding the width by 4 yds., and area is 572 sq. yds. Determine dimensions.

Let x = width in yds. and length = x + 4 yds.

Area = x(x + 4) yds.

x^2 + 4x = 572

x^2 + 4x + 2^2 = 576 **

X + 2 = sq. rt. 576 = + - 24

x = 24 - 2 = 22

x = - 24 - 2 = -26

width = 22 yds., length = 26 yds.

Are there some missing steps ?

** I don't understand the 2^2.

[Guest]

A rectangular piece of land has the length exceeding the width by 4 yds., and area is 572 sq. yds. Determine dimensions.

Let x = width in yds. and length = x + 4 yds.

Area = x(x + 4) yds.

x^2 + 4x = 572

x^2 + 4x + 2^2 = 576 **

A Quadratic can be solved by process known as completing the square
General form...if A=1 then Ax^2 +Bx = C gives x^2 + Bx = C OR x^2 + Bx - C = 0
this is equivalent to (x + B/2)^2 = C + ? .... converting it to something squared = constant + something to balance
and working out the brackets squared gives x^2 + Bx + B^2/4 = C + something
So we still have the x^2 + Bx bit, but have an extra B^2/4 that we have to add to C to keep the equation in balance
This means in the squared form we have (x + B/2)^2 = (C + (B^2/4))

In this case A=1, B=4 C=-572

Area = x(x + 4) yds.

giving our orig equation is x^2 + 4x = 572 OR x^2 + 4x - 572 = 0

Completing the square (taking half of 4) as from example above gives.............

So (x + 2)^2 = 572 + ?

which gives x^2 +4x + (4^2)/4 = 572 + ?

we see that when we write it as original equation we have an extra 4 added

So x^2 +4x + 4 = 572 + 4

x^2 + 4x was in the original equation and we now have an extra 4, so this is what we have to add this to other side to balance

So that gives us .... x^2 +4x + 4 = 576

OR in completed square form .... (x + 2)^2 = 576

Take square root of both sides to find (X + 2)

X + 2 = sq. rt. 576 = + 24 or - 24

Note:- There is a possibility of 2 answers..... x + 2 = +24 OR x + 2 = -24

for case 1..... x + 2 = +24...........

x = 24 - 2 = 22 .... so x = 22 for case 1.....

For case 2 .... x + 2 = -24 .......

x = - 24 - 2 = -26 .....so x = -26 for case 2

A minus width ( -24) is not a valid dimension so the valid one is x = 22.

width = 22 yds., length = 26 yds.

Are there some missing steps ? .........Hopefully all the missing steps added now......

** I don't understand the 2^2. ...... the 2^2 comes from squaring (x + 2)^2 or the equiv. multiplying the brackets (x + 2)(x + 2) = x*x + 2x + 2x + 2*2 brackets. ......2^2 is same as 2 x 2 is same as 2*2.

[Guest]

I am not totally clear on your steps.

[Guest]

The method used for solving the quadratic is called "completing the square"
There are lots of examples starting from the basics online.

We don't have to use this method to solve, I was only following the previous post with some more explanations of the steps.
If you post the actual bits you don't understand I will try and answer it.

It could have been solved by factorising the quadratic.....

Rearranging in standard form...... x^2 + 4x - 572 = 0

The coefficient of the x^2 term is 1 so that makes it easier....
You need to find the factors of 572 that when added or subtracted gives 4
The sign of the 572 is minus so that says the factors will be subtracted
What are the factors of 572 ..... 2 x 2 x 11 x 13 = 572 are all the prime factors and they can be combined so 2 x 11 = 22 and 2 x 13 = 26.
So also ... 22 x 26 = 572

Hindsight is a great thing, because we have already solved this in the earlier posts so we know 22 and 26 work.
But for the purpose of this example 22 and 26 work because if you subtract them you get 4.

So which bracket will I make + or -
We have to make the bracket (X + 26) and (X - 22) so that 26X - 22X gave us +4X
If you multiply out the brackets (X + 26)(X - 22) you get X^2 - 22X + 26X - 572 .... so same as original X^2 + 4X - 572

So solving by factorising ....... X^2 + 4X - 572 = 0
gives ........ (X + 26)(X - 22) = 0

So either X + 26 =0 OR X - 22 = 0
So X = -26 OR X = 22 ...... as before...

We had to make the bracket with the bigger number (X + 26) "positive sign" and (X - 22) "minus sign "so that 26X - 22X gave us positive " +4X "

[Guest]

" General form...if A=1 then Ax^2 +Bx = C gives x^2 + Bx = C OR x^2 + Bx - C = 0
this is equivalent to (x + B/2)^2 = C + ? .... converting it to something squared = constant + something to balance
and working out the brackets squared gives x^2 + Bx + B^2/4 = C + something
So we still have the x^2 + Bx bit, but have an extra B^2/4 that we have to add to C to keep the equation in balance
This means in the squared form we have (x + B/2)^2 = (C + (B^2/4)) " I don't totally understand this process.

[Guest]

Please reply. Thanks.

[Guest]

This means in the squared form we have (x + B/2)^2 = (C + (B^2/4)) " I don't totally understand this process.

If you have " X + 2 " and you want to square it, you multiply it by itself and you have (X + 2) multiplied by (X + 2).

That gives (X + 2)(X + 2) and you have to multiply each "X" and "2" in each bracket together. Some people use the FOIL notation
( Firsts, Outers, Inners, Lasts ) to make sure they follow a sequence so none are missed. so using FOIL.....gives.....
(X + 2)(X + 2) = X^2 + 2X + 2X + 4 = x^2 + 4X + 4.

You will notice that the X bit in the result is 4X and 4 is twice what we had in the brackets and since we are squaring both brackets and both are the same then in fact we could have just doubled the 2 to get 4.

This leads to another process for squaring brackets....because both are the same and contain only 2 terms ......

Square the First, plus, Square of last, plus, Twice the product of first and last. .....so here we have it....

X^2 + 2^2 + 2*2X = X^2 + 4 + 4X = X^2 + 4X + 4

So (X + 2)^2 is same as X^2 + 4X + 4 as we got before.......

Now the reverse of this procedure..............

If we start with the quadratic X^2 + 4X + 4 we can do the reverse of the above and write as a complete square eg. something in a bracket squared or by 2 brackets multiplied together to square it.

So (X + 2)^2 is same as X^2 + 4X + 4 as we got before.......
As you can see to compare it to the posted question B = 4 and the thing we put in the bracket to be squared is B/2

So ...... (x + B/2)^2 = X^2 + (B^2/4) + (2* (BX/2)) ....using this method.. "Square the First, plus, Square of last, plus, Twice the product of first and last."

So that gives us X^2 + 16/4 + 2*4X/2 = X^2 + 4 + 8X/2 = X^2 + 4 + 4X = X^2 + 4X + 4

Now if X^2 + 4X + 4 = 572 that would be fine and you could say (X + 2)^2 = 572

because....... X^2 + 4X + 4 is same as (X + 2)^2 ......

But in the original question X(X + 4) gave use X^2 + 4X .....and there was no 4 added on to the end......so it is not a complete square.
And to use "completing the square" as a method for solving we had to put the bracket (X + 2) together because 2 was half of 4 (for the 4X) but when the brackets was multiplied out we had the extra 4 so we had to add another 4 to the other side to keep balance.

So we had originally ... X^2 + 4X = 572 but we completed the square as (X + 2)^2 = 572 + ? ..... same as X^2 + 4X + 4 = 572 + 4 ...to balance.

.....................................

[Guest]

" General form...if A=1 then Ax^2 +Bx = C gives x^2 + Bx = C OR x^2 + Bx - C = 0
this is equivalent to (x + B/2)^2 = C + ? .... converting it to something squared = constant + something to balance
and working out the brackets squared gives x^2 + Bx + B^2/4 = C + something
So we still have the x^2 + Bx bit, but have an extra B^2/4 that we have to add to C to keep the equation in balance
This means in the squared form we have (x + B/2)^2 = (C + (B^2/4)) "


I didn't understand this part. I should have made that clear. Sorry about that.

[Guest]

Be spefic abot what you do not know dn not eat an elephant in one bite

[Guest]

" General form...if A=1 then Ax^2 +Bx = C gives x^2 + Bx = C OR x^2 + Bx - C = 0
this is equivalent to (x + B/2)^2 = C + ? .... converting it to something squared = constant + something to balance
and working out the brackets squared gives x^2 + Bx + B^2/4 = C + something "

I don't understand this part.

[Guest]

Break it down to the bits you dont know can you mult. Brackets. Can you coplete the square. Can you solve quadratics. And by what methods

[Guest]

'"can you mult. Brackets. Can you coplete the square. Can you solve quadratics." No.

[Guest]

Please reply. Thanks.

[Guest]

Each of these is a different topic in algebra. You need to understand the basics first and then develop what you have learned by doing harder problems after you have mastered the basics.

You should start with something like ....... http://www.algebrahelp.com/ and post basic questions to start with if you don't understand the methods.

[Guest]

Ok, I will try that. Thanks again.

[Guest]

In terms of this question the basics steps are:-...........
X is a variable thet represents the width
The length is 4 units longer......

We add on 4 that gives (X + 4) ......I put a bracket around it to keep them together...

The area is width mult. by length .... X(X + 4) ...... gives X^2 + 4X and the question says this equals 572

This gives us X^2 + 4X = 572 OR re-arranging gives X^2 + 4X - 572 = 0

The above line is in Quadratic form and the standard form is ax^2 + bx + c = 0 and we can solve for X using various methods.

The method used to solve in the earlier post was "completing the square" where we convert the original expression into a complete square, that means two brackets that when multiplied together gives us the same thing as our original expression....... Then we can take the square root of it and solve the equation........

"How completing the square works" is illustrated as below.......starting with "X + 2" (2 is half of 4) and squaring it, seeing it gives X^2 + 4X + 4 ....and how this is same as original expression with a new "4" term resulting from the multiplication which needs to be added to the other side to balance if the equation is written in (X + 2) squared form. ...................


If you have " X + 2 " and you want to square it, you multiply it by itself and you have (X + 2) multiplied by (X + 2).

That gives (X + 2)(X + 2) and you have to multiply each "X" and "2" in each bracket together. Some people use the FOIL notation
( Firsts, Outers, Inners, Lasts ) to make sure they follow a sequence so none are missed. so using FOIL.....gives.....
(X + 2)(X + 2) = X^2 + 2X + 2X + 4 = x^2 + 4X + 4.

You will notice that the X bit in the result is 4X and 4 is twice what we had in the brackets and since we are squaring both brackets and both are the same then in fact we could have just doubled the 2 to get 4.

This leads to another process for squaring brackets....because both are the same and contain only 2 terms ......

Square the First, plus, Square of last, plus, Twice the product of first and last. .....so here we have it....

X^2 + 2^2 + 2*2X = X^2 + 4 + 4X = X^2 + 4X + 4

So (X + 2)^2 is same as X^2 + 4X + 4 as we got before.......

Now the reverse of this procedure..............

If we start with the quadratic X^2 + 4X + 4 we can do the reverse of the above and write as a complete square eg. something in a bracket squared or by 2 brackets multiplied together to square it.

So (X + 2)^2 is same as X^2 + 4X + 4 as we got before.......
As you can see to compare it to the posted question B = 4 and the thing we put in the bracket to be squared is B/2

So ...... (x + B/2)^2 = X^2 + (B^2/4) + (2* (BX/2)) ....using this method.. "Square the First, plus, Square of last, plus, Twice the product of first and last."

So that gives us X^2 + 16/4 + 2*4X/2 = X^2 + 4 + 8X/2 = X^2 + 4 + 4X = X^2 + 4X + 4

Now if X^2 + 4X + 4 = 572 that would be fine and you could say (X + 2)^2 = 572

because....... X^2 + 4X + 4 is same as (X + 2)^2 ......

But in the original question X(X + 4) gave use X^2 + 4X .....and there was no 4 added on to the end......so it is not a complete square.
And to use "completing the square" as a method for solving we had to put the bracket (X + 2) together because 2 was half of 4 (for the 4X) but when the brackets was multiplied out we had the extra 4 so we had to add another 4 to the other side to keep balance.

So we had originally ... X^2 + 4X = 572 but we completed the square as (X + 2)^2 = 572 + ? ..... same as X^2 + 4X + 4 = 572 + 4 ...to balance.

[Guest]

" Square the First, plus, Square of last, plus, Twice the product of first and last. .....so here we have it.... "

X^2 + 2^2 + 2*2X = X^2 + 4 + 4X = X^2 + 4X + 4

So ...... (x + B/2)^2 = X^2 + (B^2/4) + (2* (BX/2)) ....using this method..


So that gives us X^2 + 16/4 + 2*4X/2 = X^2 + 4 + 8X/2 = X^2 + 4 + 4X = X^2 + 4X + 4 "

I still don't understand this part.

[Guest]

You need to read up and practice doing simplifying and multiplying brackets etc.
Read the earlier posts.....The example here is for multiplying (X + 2) by (X + 2) ....same as (X + 2)^2

Both brackets are the same "X" is first term in both brackets and "2" is the last term in both brackets.

When written as (X + 2)^2 The "First" term is still "X" and the second is still "2"

So do it in bits.....

Square the First .....= X^2
Square of last ..... = 2^2
Twice the product of first and last....... 2 multiplied by (2 times X)

and ADD them together....gives......

X^2 + 2^2 + 2*2X
Work out the bits that can be multiplied so as to simplify gives X^2 + 4 + 4X
Now re-arrange it into standard form ...... X^2 + 4X + 4

Now compare the equation above with the general normal form ..AX^2 + BX + C

"A" is equal to 1 so we have X^2
"B" is equal to 4 so we have 4X and B/2 is 2X
"C" is equal to 4 so we have 4

So the general form AX^2 + BX + C can be written as (X + (B/2))^2

So ...... (X + B/2)^2 = X^2 + (B^2/4) + (2* (BX/2)) ....using the method of squaring or multiplying 2 brackets shown above......
This is "Square of the first, plus, Square of the last, plus 2times the product of first and last"......

So that gives us X^2 + 16/4 + 2*4X/2 = X^2 + 4 + 8X/2 = X^2 + 4 + 4X = X^2 + 4X + 4

The above is only to let you see that (X + 2)^2 is same as X^2 + 4X + 4 ....it is nothing more complicated than that.

From studying this you should see that (X + 2)^2 is a perfect square as is (X^2 + 4X + 4) because they are the same

In the original question we had X(X + 4) which gave us (X^2 + 4X) which is not a perfect square and we could not take the square root of it to solve the equation, so we used a process of "completing the square" on the LHS of the equation using the method show above here, and we ended up with the complete square (X^2 + 4X + 4) on the LHS and you can see that there is the extra "4" so to balance the equation we had to add 4 to the RHS. But the LHS is now a complete square so we can take the square root of both sides SquareRoot(X + 2)^2 = SquareRoot(576)

So that gives X + 2 = +24 or -24 ..... so X = 24 - 2 OR X = -24 - 2

So X = 22 OR -26 .....minus 26 is not a valid width so ignore it

So the width is 22 and the length is 4 more so length is 26.

[Guest]

" You need to read up and practice doing simplifying and multiplying brackets etc.
Read the earlier posts.....The example here is for multiplying (X + 2) by (X + 2) ....same as (X + 2)^2 "

I will. Thanks again for the extra details.