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The area of one square is x 2 + 10x + 25 and the area of another squar
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14 Feb 2016, 03:15
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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
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15 Feb 2016, 15:07
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= (2x3)^2 L1= x+5 & L2= 2x3 P1 = 4( x+5) & P2=4(2x3) P1+P2=32 4( x+5) +4(2x3)=32..............> X=2 Answer: B




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Re: The area of one square is x 2 + 10x + 25 and the area of another squar
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24 Oct 2016, 10:52
x^2 + 10x + 25 = (x+5)^2 Perimeter = 4(x+5)
4x^2 − 12x + 9 = (2x3)^2 Perimeter = 4(2x3)
4(x+5)+4(2x3) = 32 x =2



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Re: The area of one square is x 2 + 10x + 25 and the area of another squar
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24 Oct 2016, 12:18
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
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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 (2x3)^2 = z(side of the seconds quared)^2 z=2x3 4y+4z = 32 y+z=8 2x3+x+5=3x2=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
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13 Mar 2017, 01:43
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(2x3) = 32 so x = 2 correct answer  B



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Re: The area of one square is x 2 + 10x + 25 and the area of another squar
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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
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28 Mar 2017, 08:09
hi vmelgargalan,
rhine29388 has correctly pointed. 4x+4y=32 4(x+5)+4(2x3)=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
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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
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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 Answer: B
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Re: The area of one square is x 2 + 10x + 25 and the area of another squar
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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^212x+9 which is (2x3) 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
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