this is quadratic word problems

## the area of a square is 165 cm greater than its perimeter. what is the length of the side of this square?

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# 4 Answers

side - x

area - x

^{2}perimeter - 4x

x

^{2}- 4x = 165x

^{2}- 4x - 165 = 0D = (-4)

^{2}+ 165*4 = 676x1 = (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+165x

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

^{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

**x=15**

Let's check our work...

x

^{2}=4x+165(15)

^{2}=[(4)(15)]+165225=60+165

225=225

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