# A rectangular patio

A rectangular patio measures 20 ft by 30 ft. By adding x feet to the width and x feet to the length, the area is doubled. Find the new dimensions of the patio.

Result

c =  30 ft
d =  40 ft

#### Solution:

$a=20 \ \text{ft} \ \\ b=30 \ \text{ft} \ \\ \ \\ (a+x)*(b+x)=2*a*b \ \\ \ \\ (20+x) \cdot \ (30+x)=2 \cdot \ 20 \cdot \ 30 \ \\ x^2 +50x -600=0 \ \\ \ \\ a=1; b=50; c=-600 \ \\ D=b^2 - 4ac=50^2 - 4\cdot 1 \cdot (-600)=4900 \ \\ D>0 \ \\ \ \\ x_{1,2}=\dfrac{ -b \pm \sqrt{ D } }{ 2a }=\dfrac{ -50 \pm \sqrt{ 4900 } }{ 2 } \ \\ x_{1,2}=\dfrac{ -50 \pm 70 }{ 2 } \ \\ x_{1,2}=-25 \pm 35 \ \\ x_{1}=10 \ \\ x_{2}=-60 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (x -10) (x +60)=0 \ \\ \ \\ c=a+x_{1}=20+10=30 \ \text{ft}$

Checkout calculation with our calculator of quadratic equations.

$d=b+x_{1}=30+10=40 \ \text{ft}$

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