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# The area of one square is x 2 + 10x + 25 and the area of another squar

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The area of one square is x 2 + 10x + 25 and the area of another squar  [#permalink]

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14 Feb 2016, 03:15
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Question Stats:

59% (02:05) correct 41% (02:11) wrong based on 374 sessions

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The area of one square is x^2 + 10x + 25 and the area of another square is 4x^2 − 12x + 9. If the sum of the perimeters of both squares is 32, what is the value of x?

A. 0
B. 2
C. 2.5
D. 4.67
E. 10

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The area of one square is x 2 + 10x + 25 and the area of another squar  [#permalink]

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15 Feb 2016, 15:07
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Bunuel wrote:
The area of one square is x^2 + 10x + 25 and the area of another square is 4x^2 − 12x + 9. If the sum of the perimeters of both squares is 32, what is the value of x?

A. 0
B. 2
C. 2.5
D. 4.67
E. 10

spotting the pattern of equations both are in form of (X+C)^2 so

A1= (x+5)^2 & A2= (2x-3)^2

L1= x+5 & L2= 2x-3

P1 = 4( x+5) & P2=4(2x-3)

P1+P2=32

4( x+5) +4(2x-3)=32..............> X=2

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Re: The area of one square is x 2 + 10x + 25 and the area of another squar  [#permalink]

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24 Oct 2016, 10:52
x^2 + 10x + 25 = (x+5)^2
Perimeter = 4(x+5)

4x^2 − 12x + 9 = (2x-3)^2
Perimeter = 4(2x-3)

4(x+5)+4(2x-3) = 32
x =2
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Re: The area of one square is x 2 + 10x + 25 and the area of another squar  [#permalink]

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24 Oct 2016, 12:18
1
Bunuel wrote:
The area of one square is x^2 + 10x + 25 and the area of another square is 4x^2 − 12x + 9. If the sum of the perimeters of both squares is 32, what is the value of x?

A. 0
B. 2
C. 2.5
D. 4.67
E. 10

$$x^2 + 10x + 25$$ = $$( x + 5 )^2$$

$$4x^2 − 12x + 9$$ = $$( 2x - 3 )^2$$

Now, $$( x + 5 ) + ( 2x - 3 ) = 32$$

Or, $$3x + 2 = 32$$

Or, $$x = 10$$

Hence , answer will be (E)

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Re: The area of one square is x 2 + 10x + 25 and the area of another squar  [#permalink]

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25 Nov 2016, 09:54
yep...agree...it's B - x=2

(x+5)^2 = Y(side of first sqare) ^2
y=x+5

(2x-3)^2 = z(side of the seconds quared)^2
z=2x-3

4y+4z = 32
y+z=8

2x-3+x+5=3x-2=8
3x=6
x=2

but 0 works too..so something is wrong with the question...

if x=0

x^2 + 10x + 25 = y^2
25 = y^2
y=5

4x^2 − 12x + 9 = z^2
9=z^2
z=3

y+z = 8

Bunuel - can you pls check, maybe instead of 0 to put 1 or 1.5? because if we plug it in - it works too..
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Re: The area of one square is x 2 + 10x + 25 and the area of another squar  [#permalink]

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13 Mar 2017, 01:43
1
Abhishek009 wrote:
Bunuel wrote:
The area of one square is x^2 + 10x + 25 and the area of another square is 4x^2 − 12x + 9. If the sum of the perimeters of both squares is 32, what is the value of x?

A. 0
B. 2
C. 2.5
D. 4.67
E. 10

$$x^2 + 10x + 25$$ = $$( x + 5 )^2$$

$$4x^2 − 12x + 9$$ = $$( 2x - 3 )^2$$

Now, $$( x + 5 ) + ( 2x - 3 ) = 32$$

Or, $$3x + 2 = 32$$

Or, $$x = 10$$

Hence , answer will be (E)

apologies for seeing a mistake
sum of perimeter of 2 squares is given as 32 whereas you are considering the sum of 2 sides of the square as 32 which would be wrong pls correct the answer
correct answer is 4 (x+5) + 4(2x-3) = 32
so x = 2
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Re: The area of one square is x 2 + 10x + 25 and the area of another squar  [#permalink]

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23 Mar 2017, 10:53
Mhmmm, yeah I used a similar approach to mvcitor. and I seems that using this method would also make option A valid. So I assume I did it wrong. But basically if we call the side of square one y, and side of square two x,

4x + 4y = 32,
x + y = 8. So we are looking for values that give us this. So we plug in the available options into the initial formulas and when x + y = 8 that is the solution.

But as Victor has said option A would also work in this case too... So I assume that getting the correct solution was out of luck
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Re: The area of one square is x 2 + 10x + 25 and the area of another squar  [#permalink]

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28 Mar 2017, 08:09
hi vmelgargalan,

rhine29388 has correctly pointed.
4x+4y=32
4(x+5)+4(2x-3)=32
Upon simplifying x=2
I guess you and mvictor are trying to equate the values of x in both squares which has different roots.

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Re: The area of one square is x 2 + 10x + 25 and the area of another squar  [#permalink]

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03 Apr 2017, 04:31
vmelgargalan wrote:
Mhmmm, yeah I used a similar approach to mvcitor. and I seems that using this method would also make option A valid. So I assume I did it wrong. But basically if we call the side of square one y, and side of square two x,

4x + 4y = 32,
x + y = 8. So we are looking for values that give us this. So we plug in the available options into the initial formulas and when x + y = 8 that is the solution.

But as Victor has said option A would also work in this case too... So I assume that getting the correct solution was out of luck

Hi,
i tried back solving , putting value of X in the equations ,
and getting A and B as a answer , like in option A put x=0 then equ.1 would be have 25 and equation 2, would have 9 then perimeter will be 5*4+3*4=32
same process for option B , we get perimeter 7*4+4*1=32
what's wrong in this can you clarify

thanks
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Re: The area of one square is x 2 + 10x + 25 and the area of another squar  [#permalink]

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06 Apr 2017, 09:31
Bunuel wrote:
The area of one square is x^2 + 10x + 25 and the area of another square is 4x^2 − 12x + 9. If the sum of the perimeters of both squares is 32, what is the value of x?

A. 0
B. 2
C. 2.5
D. 4.67
E. 10

We have to determine the side length of each square. Since the area of the first square is x^2 + 10x + 25 = (x + 5)^2, its side is x + 5. Similarly, since the area of the second square is 4x^2 − 12x + 9 = (2x - 3)^2, its side is 2x - 3. Furthermore, the perimeter of the first square is 4(x + 5) and that of the second square is 4(2x - 3). Since the sum of the perimeters of the two squares is 32, we can create the following equation:

4(x + 5) + 4(2x - 3) = 32

4x + 20 + 8x - 12 = 32

12x + 8 = 32

12x = 24

x = 2

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Re: The area of one square is x 2 + 10x + 25 and the area of another squar  [#permalink]

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26 Aug 2018, 05:58
nks2611 wrote:
vmelgargalan wrote:
Mhmmm, yeah I used a similar approach to mvcitor. and I seems that using this method would also make option A valid. So I assume I did it wrong. But basically if we call the side of square one y, and side of square two x,

4x + 4y = 32,
x + y = 8. So we are looking for values that give us this. So we plug in the available options into the initial formulas and when x + y = 8 that is the solution.

But as Victor has said option A would also work in this case too... So I assume that getting the correct solution was out of luck

Hi,
i tried back solving , putting value of X in the equations ,
and getting A and B as a answer , like in option A put x=0 then equ.1 would be have 25 and equation 2, would have 9 then perimeter will be 5*4+3*4=32
same process for option B , we get perimeter 7*4+4*1=32
what's wrong in this can you clarify

thanks

I too plugged in values in haste and marked the option as A.
Side of a square cannot be negative so 4x^2-12x+9 which is (2x-3) has to be more than than 1.5.
Re: The area of one square is x 2 + 10x + 25 and the area of another squar &nbs [#permalink] 26 Aug 2018, 05:58
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