Question

The length of the top of a table is 3 m greater than the width. The area is 70 m^2. Find the dimensions of the table.

Answer 1

Let's call the width $x$. Since the length is three more than the width, it is $x+3$.

The area (which we know to be 70) is given by the multiplication of width and length:

$x(x+3) = 70 \iff x^2+3x = 70 \iff x^2+3x-70 = 0$

To solve this equation, you can use the usual quadratic formula: given an equation $ax^2+bx+c=0$, the two solutions are

$x_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

which in your case becomes

$x_{1,2} = \frac{-3\pm\sqrt{9+280}}{2} = \frac{-3\pm 17}{2}$

So, two solutions are

$\frac{-3-17}{20} = -10,\qquad \frac{-3+17}{2} = 7$

Since we can't accept a negative length, we only accept the second solution.

So, the dimensions are 7 and 10

Answer 2

The final answer is 7 and 10! :)