Question 588867
The original area is 8*10 = 80 square inches.


The doubled area is 80*2 = 160 square inches.


The new dimensions are 8+x by 10+x and they multiply out to 160, so the equation is (8+x)(10+x) = 160

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b)


Now get the equation into standard form


(8+x)(10+x) = 160


80+18x+x^2 = 160


80+18x+x^2 - 160 = 0


x^2 + 18x - 80 = 0


Now use the quadratic formula to solve for x


x = (-b+-sqrt(b^2-4ac))/(2a)


x = (-(18)+-sqrt((18)^2-4(1)(-80)))/(2(1))


x = (-18+-sqrt(324-(-320)))/(2)


x = (-18+-sqrt(644))/2


x = (-18+sqrt(644))/2 or x = (-18-sqrt(644))/2


x = (-18+2*sqrt(161))/2 or x = (-18-2*sqrt(161))/2


and approximate the solutions above to get


x = 3.6885775 or x = -21.6885775


toss out the negative solution to get 


x = 3.6885775


So you should add about 3.6885775 inches to each dimension to double the area of the photograph.


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C)

Add 8 and 3.6885775 to get 11.6885775. This is the new width


Add 10 and 3.6885775 to get 13.6885775. This is the new length