Question

An animal welfare trust decides to build an enclosure having floor area $480{\text{ }}{{\text{m}}^2}$ for injured or abandoned dogs and cattle. If the length is 4m more than thrice its width, construct the quadratic equation to find the length and width. What value do you imbibe from this act?

Answer 1

Hint: It is given that an animal trust decides to build an enclosure. Enclosure is an area that is surrounded by a barrier. Hence we can clearly say that the enclosure must be the form of a rectangle. So we can use formulae or properties of rectangles.

Formula used: Area of rectangle = ${length \times breadth}$
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step-by-step answer:
First, we are going to assume that the breadth of the enclosure is $x$. So, ${\text{breadth = }}x$
It is also given that the length is 4m more than thrice its width.
Hence we have, ${\text{length = 3}}x{\text{ + 4}}$
Now we have the length and breadth of the enclosure that is rectangle.
It is also given that the area of the enclosure is $480{\text{ }}{{\text{m}}^2}$
Now we are going to consider the formula of area of the enclosure, that is, rectangle.
Area of enclosure = ${length \times breadth}$ --(ii)
Now we are going to substitute the values on the above equation. Then we get,
 $x(3x + 4) = 480$
Now we are going to simplify the above equation.
$3{x^2} + 4x = 480$
Now we are going to subtract 480 on both sides so that the right hand side will be zero.
$3{x^2} + 4x - 480 = 0$ ----(iii)
We already know that any equation of the form $a{x^2} + bx + c = 0$ is called a quadratic equation.
So, $3{x^2} + 4x - 480 = 0$ is a quadratic equation where $a = 3$, $b = 4$ and $c = - 480$
Now we are going to solve the above quadratic equation to get the value of the $x$
To solve for $x$, we are going to use the quadratic formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Now we are going to substitute the values of a, b and c on the above formula. Then we get,
$x = \dfrac{{ - 4 \pm \sqrt {{4^2} - 4 \times 3 \times ( - 430)} }}{{2 \times 3}}$
We are going to simplify the above equation.
$x = \dfrac{{ - 4 \pm \sqrt {16 + 12 \times 480} }}{6}$
Now we are going to use the BODMAS rule to simplify the term that is in the square root. First multiplication holds and then the addition holds.
$x = \dfrac{{ - 4 \pm \sqrt {16 + (12 \times 480)} }}{6}$
First we multiply, $12 \times 480$,
$x = \dfrac{{ - 4 \pm \sqrt {16 + 5760} }}{6}$
Then we add, 16+5760,
$x = \dfrac{{ - 4 \pm \sqrt {5776} }}{6}$
Now we are going to substitute $\sqrt {5776} = 76$ on the above equation. Then we get,
$x = \dfrac{{ - 4 \pm 76}}{6}$
Length and breadth are never taken as a negative value. So, we can skip the negative operation of the above equation and we are going to consider only addition operations.
\[x = \dfrac{{ - 4 + 76}}{6}\]
\[x = \dfrac{{ - 4 + 76}}{6}\]
\[x = \dfrac{{72}}{6}\]
We are going to simplify the above equation to get the value of \[x\]
Therefore, \[x = 12\]
Hence the breadth of the enclosure, \[x = 12{\text{ m}}\]
Now we are going to find the length of the enclosure.
It is already given that, ${\text{length = 3}}x{\text{ + 4}}$
We are going to substitute \[x = 12{\text{ m}}\]
Length of the closure = $3x + 4$ = $3 \times 12 + 4$
Now we are going to use the BODMAS rule to solve the above equation.
Hence, ${\text{length = }}36 + 4 = 40{\text{ m}}$
So, the length of the enclosure is ${\text{40 m}}$ and the breadth of the enclosure is \[{\text{12 m}}\].

Note: One must understand the nature of these types of questions. For example, in this problem we have to find out what enclosure is and how it looks like. Without finding the structure of the enclosure, we can’t proceed with this problem. If we are able to find what enclosure (area that is of the form of a rectangle) is we will easily be able to proceed and solve the problem using its formula or property.