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

Question

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

Mo2men · Accepted Answer

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

Answer: B

law258 · Answer

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

Abhishek009 · Answer

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)

mvictor · Answer

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..

rhine29388 · Answer

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 
correct answer - B

vmelgargalan · Answer

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

sarugiri · Answer

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.

nks2611 · Answer

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 :shock:

same process for option B , we get perimeter 7*4+4*1=32 :oops:

what's wrong in this can you clarify thanks

JeffTargetTestPrep · Answer

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

isasda66 · Answer

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.

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

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