# Rectangle field

The field has a shape of a rectangle having a length of 119 m and a width of 19 m. , How many meters have to shorten its length and increase its width to maintain its area and circumference increased by 24 m?

Correct result:

x =  102 m
y =  114 m

#### Solution:

$119 \cdot \ 19=(119-x) \cdot \ (19+y) \ \\ 2(119+19) + 24=2 \cdot \ ((119-x)+(19+y)) \ \\ \ \\ y=x+12 \ \\ \ \\ 119*19=(119-x)*(19+(x+12)) \ \\ \ \\ 119 \cdot \ 19=(119-x) \cdot \ (19+(x+12)) \ \\ x^2 -88x -1428=0 \ \\ \ \\ a=1; b=-88; c=-1428 \ \\ D=b^2 - 4ac=88^2 - 4\cdot 1 \cdot (-1428)=13456 \ \\ D>0 \ \\ \ \\ x_{1,2}=\dfrac{ -b \pm \sqrt{ D } }{ 2a }=\dfrac{ 88 \pm \sqrt{ 13456 } }{ 2 } \ \\ x_{1,2}=\dfrac{ 88 \pm 116 }{ 2 } \ \\ x_{1,2}=44 \pm 58 \ \\ x_{1}=102 \ \\ x_{2}=-14 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (x -102) (x +14)=0 \ \\ \ \\ x=x_{1}=102 \ \text{m}$

Checkout calculation with our calculator of quadratic equations.

$y=x+12=102+12=114 \ \\ a=119-x=119-102=17 \ \text{m} \ \\ b=19+y=19+114=133 \ \text{m} \ \\ \ \\ y=114 \ \text{m}$

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