Ken has a rectangular shaped vegetable garden that is 15 feet by 9 feet. He would like to increase the length and width by the same amount, x, so the new garden is twice the current area. Find the value of x to the nearest tenth of a foot feet.

Answer :

The current area is 15 x 9 = 135 square feet.

He wants to increase both the length and width by X:

Set up an equation:

(15 +x) * (9 +x) = 135 * 2

Simplify :

x^2 + 24x  +  135 = 270

Subtract 270 from both sides:

x^2 + 24x - 135 = 0

Use the quadratic formula to solve for x:

-24 +/- √(24^2 - 4(1*-135) / 2*1

x = 4.7 or -28.7

The answer has to be a positive value, so x = 4.7 feet.

The length and width of the rectangular shaped vegetable garden will be increased in 4.7 ft

Quadrilaterals

There are different types of quadrilaterals, for example: square, rectangle, rhombus, trapezoid and parallelogram.  Each type is defined accordingly to its length of sides and angles. For example, in a rectangle,  the opposite sides are equal and parallel and their interior angles are equal to 90°.

The area of rectangles can be found from the equation:

A=length(l) x height (h).

Thus, initial area is Ao=lh= 15*9=135 ft².

After that, you should convert the information given in the title of the question in an algebraic expression. Hence,

                                   (15+x)*(9+x)=2Ao

                                 (15+x)*(9+x)=2*135

                                  135+15x+9x+x²=270

                                    x²+24x-135=0

Solving the quadratic formula.

                             [tex]x_{1,\:2}=\frac{-24\pm \sqrt{24^2-4\cdot \:1\cdot \left(-135\right)}}{2\cdot \:1}[/tex]

                            [tex]x_{1,\:2}=\frac{-24\pm \:6\sqrt{31}}{2\cdot \:1}=-12\pm \:3\sqrt{31}}[/tex]

Thus, [tex]x_{1}=-12+3\sqrt{31}}=4.7[/tex] and [tex]x_{2}=-12-3\sqrt{31}}=-28.7[/tex]. The value of x is equal to 4.7 ft since  x must be positive number.

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