UpStudy Free Solution:

To determine the dimensions of the base of the doghouse, we need to set up an equation based on the given information:

1. The area of the rectangular base is 21 square feet.

2. The length (\(L\)) is 4 feet more than the width (\(x\)).

We can express the length in terms of the width:

\[L = x + 4\]

The area of the rectangle is given by the product of the length and the width:

\[\text { Area} = \text { Length} \times \text { Width} \]

Substitute the given values into this formula:

\[21 = ( x + 4) \times x\]

This equation can be simplified to:

\[21 = x( x + 4) \]

Expanding the right-hand side gives:

\[21 = x^ 2 + 4x\]

Rearranging the equation to standard quadratic form, we get:

\[x^ 2 + 4x - 21 = 0\]

So, the equation that can be used to determine the possible dimensions of the base of the doghouse is:

\[x^ 2 + 4x - 21 = 0\]

To find the possible dimensions, you would solve this quadratic equation for \(x\).

Supplemental Knowledge

Quadratic equations are polynomial equations of the form:

\[ax^ 2 + bx + c = 0\]

where \(a\), \(b\), and \(c\) are constants, and \(x\) represents the variable. These equations can be solved using various methods such as factoring, completing the square, or applying the quadratic formula:

\[x = \frac { - b \pm \sqrt { b^ 2 - 4ac} } { 2a} \]

In this context, we have a quadratic equation derived from a problem involving the area of a rectangle. The steps to solve it typically involve:

1. Identifying coefficients: Recognize the values of \(a\), \(b\), and \(c\) in your equation.

2. Applying the quadratic formula: Substitute these values into the formula to find the solutions for \(x\).

3. Interpreting solutions: Since we're dealing with physical dimensions (length and width), only positive solutions make sense.

For example, given:

\[x^ 2 + 4x - 21 = 0\]

We identify:

- \(a = 1\)

- \(b = 4\)

- \(c = - 21\)

Using the quadratic formula:

\[x = \frac { - 4 \pm \sqrt { 4^ 2 - 4( 1) ( - 21) } } { 2( 1) } = \frac { - 4 \pm \sqrt { 16 + 84} } { 2} = \frac { - 4 \pm 10} { 2} \]

This yields two potential solutions for \(x\):

\[x = 3.0,\space - 7.0\]

Since dimensions cannot be negative, we discard \(x = - 7.0\). Thus, the width is \(x = 3.0\) feet and length is:

\[L = x + 4 = 3 + 4= 7\] feet.

Quadratic equations are essential in solving many real-world issues in mathematics and beyond, which makes UpStudy your go-to resource for easy quadratic education! From algebraic challenges to advanced topics in algebra, UpStudy makes learning simple and enjoyable! Our platform connects you with elite tutors available 24/7 so that you can quickly master any subject effortlessly - join millions who trust UpStudy with their academic success - your ultimate partner when facing homework challenges!