+101 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?
+9481 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)
3s2 = s2 + 9s + 18
Subtract 3s2 from both sides.
0 = -2s2 + 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.
\(s=6\text{ cm}\)
+9481 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)
3s2 = s2 + 9s + 18
Subtract 3s2 from both sides.
0 = -2s2 + 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.
\(s=6\text{ cm}\)
