^{2}

^{2}- 4x = 165

^{2}- 4x - 165 = 0

^{2}+ 165*4 = 676

this is quadratic word problems

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side - x

area - x^{2}

perimeter - 4x

x^{2} - 4x = 165

x^{2} - 4x - 165 = 0

D = (-4)^{2} + 165*4 = 676

x1 = (4 + 26) / 2 = 15

x2 = (4 - 26) / 2 = -11

Disregard x2 since it is negative

Answer is : side of the square is 15cm.

Hey Raven -- an intuitive approach ... perimeter 4d must be an even # ...

area d*d seems to be an odd multiple of 5 to produce an "offset" of 165 ... 5x5 too small

... try 15x15 ==> 225 minus 60 is 165 ==> **d= 15cm** ... Best regards :)

Hi Raven;

perimeter=4x

area=x^{2}

x^{2}=4x+165

x^{2}-4x-165=0

For the FOIL...

FIRST must be (x)(x)=x^{2}

OUTER and INNER must add-up to -4x.

LAST must be (11)(15) or (15)(11) and one number must be negative to render the product of -165.

(x+11)(x-15)=0

Let's FOIL...

FIRST...(x)(x)=x^{2}

OUTER...(x)(-15)=-15x

INNER...(11)(x)=11x

LAST...(11)(-15)=-165

x^{2}-15x+11x-165=0

x^{2}-4x-165=0

(x+11)(x-15)=0

Either parenthetical equation must equal zero...

x+11=0

x=-11--not applicable. Measurements cannot be negative.

x-15=0

Let's check our work...

x^{2}=4x+165

(15)^{2}=[(4)(15)]+165

225=60+165

225=225

Tom D. | Very patient Math Expert who likes to teachVery patient Math Expert who likes to te...

I don't like the way this question is phrased because it is mixing an area (cm^2) with the linear dimension of perimeter (cm). That is never good practice. We'll assume the question addresses the magnitude of the area (cm^2) with the perimeter (cm).

s^2 = 4s + 165

s^2 - 4s - 165 =0

(S+11)(S-15) =0 --->S=15 cm

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