Certainly, let's solve each part of the problem step by step.
(a) Show that [tex]\( 2x^2 + 5x - 58 = 0 \)[/tex]
We are given the dimensions of the base of an open rectangular box. The length of the base is [tex]\((2x + 5)\)[/tex] cm and the width is [tex]\(x\)[/tex] cm. The area of the base is given as [tex]\(58\)[/tex] cm².
The area of a rectangle is given by the product of its length and width. So, we have:
[tex]\[
\text{Area} = \text{Length} \times \text{Width}
\][/tex]
Substituting the given values:
[tex]\[
58 = (2x + 5) \times x
\][/tex]
Simplifying, we get:
[tex]\[
58 = 2x^2 + 5x
\][/tex]
Rearrange the equation to standard quadratic form [tex]\(ax^2 + bx + c = 0\)[/tex]:
[tex]\[
2x^2 + 5x - 58 = 0
\][/tex]
Thus, we have shown that:
[tex]\[
2x^2 + 5x - 58 = 0
\][/tex]
(b) (i) Solve the equation [tex]\(2x^2 + 5x - 58 = 0\)[/tex]
To solve the quadratic equation [tex]\(2x^2 + 5x - 58 = 0\)[/tex], we can use the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
where [tex]\(a = 2\)[/tex], [tex]\(b = 5\)[/tex], and [tex]\(c = -58\)[/tex].
Substitute these values into the formula:
[tex]\[
x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 2 \cdot (-58)}}{2 \cdot 2}
\][/tex]
[tex]\[
x = \frac{-5 \pm \sqrt{25 + 464}}{4}
\][/tex]
[tex]\[
x = \frac{-5 \pm \sqrt{489}}{4}
\][/tex]
We get two solutions:
[tex]\[
x = \frac{-5 + \sqrt{489}}{4} \approx 4.28 \quad (\text{correct to two decimal places})
\][/tex]
[tex]\[
x = \frac{-5 - \sqrt{489}}{4} \approx -6.78 \quad (\text{correct to two decimal places})
\][/tex]
Since a width of a box cannot be negative, we discard the negative solution.
Thus, the valid solution is:
[tex]\[
x \approx 4.28
\][/tex]
(b) (ii) Hence calculate the volume of the box, stating the units of your answer.
The height of the open box is given by [tex]\((x - 2)\)[/tex] cm. We already found [tex]\(x \approx 4.28\)[/tex].
So the height is:
[tex]\[
\text{Height} = 4.28 - 2 \approx 2.28 \quad (\text{correct to two decimal places})
\][/tex]
The volume of the box is given by the product of its base area and height:
[tex]\[
\text{Volume} = \text{Length} \times \text{Width} \times \text{Height}
\][/tex]
We know:
[tex]\[
\text{Length} = 2x + 5 = 2(4.28) + 5 \approx 13.56 \quad (\text{correct to two decimal places})
\][/tex]
[tex]\[
\text{Width} = x \approx 4.28
\][/tex]
[tex]\[
\text{Height} = x - 2 \approx 2.28
\][/tex]
Substituting these values, we get:
[tex]\[
\text{Volume} \approx 13.56 \times 4.28 \times 2.28 \approx 132.14 \quad (\text{correct to two decimal places})
\][/tex]
Thus, the volume of the box is approximately:
[tex]\[
132.14 \text{ cm}^3
\][/tex]
