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Maths question - Solving quadratics

The question is: The diagram shows two joined rectangles. The total area of the compound shape ABCDEF = 36cm^2 (squared) . By considering the areas of the two rectangles, show that 2x^2 -5x -18 = 0 and hence find the value of the length AB.
Note: Please make sure your final line only shows your final answer written as "AB = .."
A screenshot of the question is : https://prnt.sc/26jrzhq

There is a question on mathswatch which I definitely have the correct answer- but it only gives me 3/6 marks, so is it something to do with how I formatted the answer, or did I do something wrong in the working out? Can any of you help me spot the problem? The answer which I entered is:

AB = 9/2
given that In rectangle DEF
DE = (x-2) cm
EF = CD+AB
EF =2x+x
EF = 3x
Area of DEF rectangle = (x-2) (3x)
DEF = (x-2) (3x)
= 3x²-6x equation 1
In rectangle ABC
AB=x cm
CB =(x-4) cm
Area of ABC = (x)(x-4)
ABC²= x²-4x
ABCDEF² = 36cm²
36= 3x²-6x+ x²-4x
36= 4x²-10x
4x²-10x-36=0
4x² -18x + 8x - 36 = 0
4x(x + 2)-18(x + 2) = 0
(4x - 18)(x + 2) = 0
x = 18/4 or -2
x = -2
x = 18/4 or x = -2
dividing by 2
(4x²-10x-36=0)/2 = 2x²-5x-18=0
=2x²-5x-18=0 proved
2x²+4x-9x-18=0
2x(x+2)-9(x+2) =0
2x-9 =0
x+2=0
x=9/2
x=-2
side never be negative so we take positive value of
AB = 9/2
AB = 4.5
x = 9/2
Original post by krugers123
The question is: The diagram shows two joined rectangles. The total area of the compound shape ABCDEF = 36cm^2 (squared) . By considering the areas of the two rectangles, show that 2x^2 -5x -18 = 0 and hence find the value of the length AB.
Note: Please make sure your final line only shows your final answer written as "AB = .."
A screenshot of the question is : https://prnt.sc/26jrzhq

There is a question on mathswatch which I definitely have the correct answer- but it only gives me 3/6 marks, so is it something to do with how I formatted the answer, or did I do something wrong in the working out? Can any of you help me spot the problem? The answer which I entered is:

AB = 9/2
given that In rectangle DEF
DE = (x-2) cm
EF = CD+AB
EF =2x+x
EF = 3x
Area of DEF rectangle = (x-2) (3x)
DEF = (x-2) (3x)
= 3x²-6x equation 1
In rectangle ABC
AB=x cm
CB =(x-4) cm
Area of ABC = (x)(x-4)
ABC²= x²-4x
ABCDEF² = 36cm²
36= 3x²-6x+ x²-4x
36= 4x²-10x
4x²-10x-36=0
4x² -18x + 8x - 36 = 0
4x(x + 2)-18(x + 2) = 0
(4x - 18)(x + 2) = 0
x = 18/4 or -2
x = -2
x = 18/4 or x = -2
dividing by 2
(4x²-10x-36=0)/2 = 2x²-5x-18=0
=2x²-5x-18=0 proved
2x²+4x-9x-18=0
2x(x+2)-9(x+2) =0
2x-9 =0
x+2=0
x=9/2
x=-2
side never be negative so we take positive value of
AB = 9/2
AB = 4.5
x = 9/2

I will highlight bits I don't like:

given that In rectangle DEF [this is not four letters and it needs to be]
DE = (x-2) cm
[EF = CD+AB
EF =2x+x] not really needed
EF = 3x
Area of rectangle = (x-2) (3x) [three letters aren't a rectangle
DEF = (x-2) (3x) ..... write 3x(x - 2)
= 3x²-6x equation 1 ... delete equation 1 words
In rectangle ABC [again this is three letters - call it something else]
AB=x cm
CB =(x-4) cm
Area of ABC = (x)(x-4) .... write as x(x - 4)
ABC²= x²-4x ...... left square .. what is the power of 2 doing on the left?
ABCDEF² = 36cm² ... write Whole area
36= 3x²-6x+ x²-4x
36= 4x²-10x
4x²-10x-36=0 ... this is';t what you had to prove ... halve everything
4x² -18x + 8x - 36 = 0 ... delete all the rest
4x(x + 2)-18(x + 2) = 0
(4x - 18)(x + 2) = 0
x = 18/4 or -2
x = -2
x = 18/4 or x = -2
dividing by 2
(4x²-10x-36=0)/2 = 2x²-5x-18=0
=2x²-5x-18=0 proved .... all these equal signs are not needed
2x²+4x-9x-18=0
2x(x+2)-9(x+2) =0 no show factorised form then either x + 2 = 0 or .... rewrite
2x-9 =0
x+2=0
x=9/2
x=-2
side never be negative so we take positive value of
AB = 9/2
AB = 4.5
x = 9/2 not in correct form


EDIT it down as i suggest but Mathswatch is notoriously fussy
Original post by Muttley79
I will highlight bits I don't like:

given that In rectangle DEF [this is not four letters and it needs to be]
DE = (x-2) cm
[EF = CD+AB
EF =2x+x] not really needed
EF = 3x
Area of rectangle = (x-2) (3x) [three letters aren't a rectangle
DEF = (x-2) (3x) ..... write 3x(x - 2)
= 3x²-6x equation 1 ... delete equation 1 words
In rectangle ABC [again this is three letters - call it something else]
AB=x cm
CB =(x-4) cm
Area of ABC = (x)(x-4) .... write as x(x - 4)
ABC²= x²-4x ...... left square .. what is the power of 2 doing on the left?
ABCDEF² = 36cm² ... write Whole area
36= 3x²-6x+ x²-4x
36= 4x²-10x
4x²-10x-36=0 ... this is';t what you had to prove ... halve everything
4x² -18x + 8x - 36 = 0 ... delete all the rest
4x(x + 2)-18(x + 2) = 0
(4x - 18)(x + 2) = 0
x = 18/4 or -2
x = -2
x = 18/4 or x = -2
dividing by 2
(4x²-10x-36=0)/2 = 2x²-5x-18=0
=2x²-5x-18=0 proved .... all these equal signs are not needed
2x²+4x-9x-18=0
2x(x+2)-9(x+2) =0 no show factorised form then either x + 2 = 0 or .... rewrite
2x-9 =0
x+2=0
x=9/2
x=-2
side never be negative so we take positive value of
AB = 9/2
AB = 4.5
x = 9/2 not in correct form


EDIT it down as i suggest but Mathswatch is notoriously fussy

I only saw you highlight the "DEF" part, is that the only text I need to remove?
Original post by krugers123
I only saw you highlight the "DEF" part, is that the only text I need to remove?

No - highlighting got too awkward - read it through - far too much in the wrong order.
Original post by Muttley79
No - highlighting got too awkward - read it through - far too much in the wrong order.

Ok, but will that guarantee the other 3 marks? Doesn't mathswatch's API just scan the textarea and mark as to whether the correct things are included?
Original post by krugers123
Ok, but will that guarantee the other 3 marks? Doesn't mathswatch's API just scan the textarea and mark as to whether the correct things are included?

No - you've solved it before you got the equation it wanted
Original post by krugers123
Ok, but will that guarantee the other 3 marks? Doesn't mathswatch's API just scan the textarea and mark as to whether the correct things are included?

Have you sorted this?
Original post by Muttley79
Have you sorted this?

Yes, thank you

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