The formula for the area of a rectangle is [latex]\text{Area}=\text{length}\cdot\text{width}[/latex], or [latex]A=l\cdot{w}[/latex].

[latex]\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,A=l\cdot{w}\\\,\text{width}=w\\\text{length}=w+7\\\,\,\,\,\,\text{area}=30\\\\30=\left(w+7\right)\left(w\right)\end{array}[/latex]

Multiply.

[latex]30=w^{2}+7w[/latex]

Subtract [latex]30[/latex] from both sides to set the equation equal to [latex]0[/latex].

[latex]w^{2}+7w–30=0[/latex]

Find two numbers whose product is [latex]−30[/latex] and whose sum is [latex]7[/latex], and write the middle term as [latex]10w–3w[/latex].

[latex]w^{2}+10w–3w–30=0[/latex]

Factor w out of the first pair and [latex]−3[/latex] out of the second pair.

[latex]w\left(w+10\right)-3\left(w+10\right)=0[/latex]

Factor out [latex]w+10[/latex].

[latex]\left(w–3\right)\left(w+10\right)=0[/latex]

Use the Zero Product Property to solve for w.

[latex]\begin{array}{l}w-3=0\,\,\,\,\,\,\text{or}\,\,\,\,\,\,w+10=0\\\,\,\,\,\,\,\,w=3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,w=-10\end{array}[/latex]

The solution [latex]w=−10[/latex] does not work for this application, as the width cannot be a negative number, we discard the [latex]−10[/latex]. So, the width is [latex]3[/latex] feet.

The width = [latex]3[/latex] feet

Substitute [latex]w=3[/latex] into the expression [latex]w+7[/latex] to find the length: [latex]3+7=10[/latex].

The length is [latex]3+7=10[/latex] feet

Answer

The width of the garden is [latex]3[/latex] feet, and the length is [latex]10[/latex] feet.

When the ball hits the ground, the height is [latex]0[/latex]. Substitute [latex]0[/latex] for h.

[latex]\begin{array}{c}h=-16t^{2}-10t+200\\0=-16t^{2}-10t+200\\-16t^{2}-10t+200=0\end{array}[/latex]

This equation is difficult to solve by factoring or by completing the square, so solve it by applying the Quadratic Formula, [latex]x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}[/latex]. In this case, the variable is t rather than x. [latex]a=−16,b=−10[/latex], and [latex]c=200[/latex].

[latex]t=\frac{-(-10)\pm \sqrt{{{(-10)}^{2}}-4(-16)(200)}}{2(-16)}[/latex]

Simplify. Be very careful with the signs.

[latex]\begin{array}{l}t=\frac{10\pm \sqrt{100+12800}}{-32}\\\,\,=\frac{10\pm \sqrt{12900}}{-32}\end{array}[/latex]

Use a calculator to find both roots.

t is approximately [latex]−3.86[/latex] or [latex]3.24[/latex].

Consider the roots logically. One solution, [latex]−3.86[/latex], cannot be the time because it is a negative number. The other solution, [latex]3.24[/latex] seconds, must be when the ball hits the ground.

The ball hits the ground approximately [latex]3.24[/latex] seconds after being thrown.

Sketch the problem. Since you do not know the width of the border, you will let the variable [latex]x[/latex] represent the width.

In the diagram. The original quilt is indicated by the red rectangle. The border is the area between the red and blue lines.

Since each side of the original [latex]4[/latex] by [latex]5[/latex] quilt has the border of width x added, the length of the quilt with the border will be [latex]5+2x[/latex], and the width will be [latex]4+2x[/latex].

(Both dimensions are written in terms of the same variable, and you will multiply them to get an area! This is where you might start to think that a quadratic equation might be used to solve this problem.)

You are only interested in the area of the border strips. Write an expression for the area of the border.

Area of border = Area of the blue rectangle minus the area of the red rectangle

Area of border[latex]=\left(4+2x\right)\left(5+2x\right)–\left(4\right)\left(5\right)[/latex]

There are [latex]10[/latex] sq ft of fabric for the border, so set the area of border to be [latex]10[/latex].

[latex]10=\left(4+2x\right)\left(5+2x\right)–20[/latex]

Multiply [latex]\left(4+2x\right)\left(5+2x\right)[/latex].

[latex]10=20+8x+10x+4x^{2}–20[/latex]

Simplify.

[latex]10=18x+4x^{2}[/latex]

Subtract [latex]10[/latex] from both sides so that you have a quadratic equation in standard form and can apply the Quadratic Formula to find the roots of the equation.

[latex]\begin{array}{c}0=18x+4x^{2}-10\\\\\text{or}\\\\4x^{2}+18x-10=0\\\\2\left(2x^{2}+9x-5\right)=0\end{array}[/latex]

Factor out the greatest common factor, [latex]2[/latex], so that you can work with the simpler equivalent equation, [latex]2x^{2}+9x–5=0[/latex].

[latex]\begin{array}{r}2\left(2x^{2}+9x-5\right)=0\\\\\frac{2\left(2x^{2}+9x-5\right)}{2}=\frac{0}{2}\\\\2x^{2}+9x-5=0\end{array}[/latex]

Use the Quadratic Formula. In this case, [latex]a=2,b=9[/latex], and [latex]c=−5[/latex].

[latex]\begin{array}{l}x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\\\x=\frac{-9\pm \sqrt{{{9}^{2}}-4(2)(-5)}}{2(2)}\end{array}[/latex]

Simplify.

[latex]x=\frac{-9\pm \sqrt{121}}{4}=\frac{-9\pm 11}{4}[/latex]

Find the solutions, making sure that the [latex]\pm[/latex] is evaluated for both values.

[latex]\begin{array}{c}x=\frac{-9+11}{4}=\frac{2}{4}=\frac{1}{2}=0.5\\\\\text{or}\\\\x=\frac{-9-11}{4}=\frac{-20}{4}=-5\end{array}[/latex]

Ignore the solution [latex]x=−5[/latex], since the width could not be negative.

The width of the border should be [latex]0.5[/latex] ft.