The area of a square can be tripled by increasing its length by 6 centimeters and increasing its width by 3 centimeters. What is the length of the side of the square?

Whiz333
Apr 15, 2017

#1**+2 **

We can make these equations from the info in the question:

a = s * s

3a = (s + 6) * (s + 3)

Substitute.

3(s * s) = (s + 6) * (s + 3)

3s^{2} = s^{2} + 9s + 18

Subtract 3s^{2} from both sides.

0 = -2s^{2} + 9s + 18

Now we can use the quadratic formula to solve for s.

\(s = {-9 \pm \sqrt{9^2-4(-2)(18)} \over 2(-2)} \\~\\ s= {-9 \pm \sqrt{81+144} \over -4} \\~\\ s= {-9 \pm \sqrt{225} \over -4} \\~\\ s= \frac{-9\pm15}{-4} \\~\\s =\frac{ -9 + 15} { -4}=-\frac32\text{ or } s=\frac{-9-15}{-4}=6\)

We know the side length is positive.

\(s=6\text{ cm}\)

hectictar
Apr 15, 2017

#1**+2 **

Best Answer

We can make these equations from the info in the question:

a = s * s

3a = (s + 6) * (s + 3)

Substitute.

3(s * s) = (s + 6) * (s + 3)

3s^{2} = s^{2} + 9s + 18

Subtract 3s^{2} from both sides.

0 = -2s^{2} + 9s + 18

Now we can use the quadratic formula to solve for s.

\(s = {-9 \pm \sqrt{9^2-4(-2)(18)} \over 2(-2)} \\~\\ s= {-9 \pm \sqrt{81+144} \over -4} \\~\\ s= {-9 \pm \sqrt{225} \over -4} \\~\\ s= \frac{-9\pm15}{-4} \\~\\s =\frac{ -9 + 15} { -4}=-\frac32\text{ or } s=\frac{-9-15}{-4}=6\)

We know the side length is positive.

\(s=6\text{ cm}\)

hectictar
Apr 15, 2017