d. A square piece of land has an area of x2 square metres. A smaller square of area 25 square metres is…

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d. A square piece of land has an area of x2 square metres. A smaller square of area 25 square metres is removed from one corner. Write an expression for the remaining area and factorise it. (3 marks)

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d. A square piece of land has an area of $x^2$ square metres. A smaller square of area 25 square metres is removed from one corner. Write an expression for the remaining area and factorise it. (3 marks)

Write the expression for the remaining area.
The remaining area is the original area minus the removed area.
$\text{Remaining Area} = x^2 - 25$
Factorise the expression using the difference of two squares formula, $A^2 - B^2 = (A - B)(A + B)$ .
Here, $A=x$ and $B=\sqrt{25}=5$ .
$x^2 - 5^2 = (x - 5)(x + 5)$
The expression for the remaining area is $\boxed{x^2 - 25}$ and its factorised form is $\boxed{(x - 5)(x + 5)}$ .

3. a. Simplify the following (4 marks)
- i. $\frac{x}{x+1} - \frac{2x}{x+2}$
Find a common denominator, which is $(x+1)(x+2)$ .
$$ \frac{x(x+2)}{(x+1)(x+2)} - \frac{2x(x+1)}{(x+1)(x+2)} $$

Combine the fractions and expand the numerator.

$$
\frac{x(x+2) - 2x(x+1)}{(x+1)(x+2)} = \frac{x^2 + 2x - (2x^2 + 2x)}{(x+1)(x+2)}
$$

Simplify the numerator.
$$ \frac{x^2 + 2x - 2x^2 - 2x}{(x+1)(x+2)} = \frac{-x^2}{(x+1)(x+2)} $$ The simplified expression is $\boxed{\frac{-x^2}{(x+1)(x+2)}}$.
- ii. $\frac{4m^2 - 9}{(m-1)^2} \div \frac{2m+3}{m^2-m}$

Factorise the numerators and denominators.

$4m^2 - 9 = (2m - 3)(2m + 3)$ (difference of two squares)  
$m^2 - m = m(m - 1)$  
The expression becomes:
$$
\frac{(2m - 3)(2m + 3)}{(m-1)^2} \div \frac{2m+3}{m(m-1)}
$$

Change division to multiplication by the reciprocal of the second fraction.
$$ \frac{(2m - 3)(2m + 3)}{(m-1)^2} \times \frac{m(m-1)}{2m+3} $$
Cancel common factors $(2m+3)$ and $(m-1)$ .
$$ \frac{(2m - 3)\cancel{(2m + 3)}}{(m-1)\cancel{(m-1)}} \times \frac{m\cancel{(m-1)}}{\cancel{2m+3}} = \frac{m(2m - 3)}{m-1} $$ The simplified expression is $\boxed{\frac{m(2m - 3)}{m-1}}$.
3. b. For the fraction $\frac{(x-2)(x+3)}{x^2-4}$
- i. Find the value(s) of $x$ that make the fraction undefined. Explain your reasoning. (3 marks)
A fraction is undefined when its denominator is equal to zero.
Set the denominator to zero: $$ x^2 - 4 = 0 $$

Solve for $x$ .

$$
x^2 = 4
$$
$$
x = \pm\sqrt{4}
$$
$$
x = \pm 2
$$
The values of $x$ that make the fraction undefined are $\boxed{x = 2 \text{ and } x = -2}$. This is because these values make the denominator zero, which results in division by zero.

ii. Find the value(s) of $x$ that make the fraction zero. Explain your reasoning. (3 marks)

A fraction is zero when its numerator is equal to zero, provided the denominator is not also zero for that value of $x$ .
Set the numerator to zero: $$ (x-2)(x+3) = 0 $$

Solve for $x$ .

$$
x-2 = 0 \implies x = 2
$$
$$
x+3 = 0 \implies x = -3
$$

Check if these values make the denominator zero.
For $x=2$, the denominator $x^2-4 = 2^2-4 = 4-4=0$. Since the denominator is zero, $x=2$ does not make the fraction zero; it makes it undefined. For $x=-3$, the denominator $x^2-4 = (-3)^2-4 = 9-4=5$. The denominator is not zero. The value of $x$ that makes the fraction zero is $\boxed{x = -3}$. This is because when $x=-3$, the numerator is zero and the denominator is non-zero.
3. c. A company's profit is modelled by the fraction $\frac{2x}{x-5}$ where $x$ represents units sold.
- i. Explain what happens to the profit when $x=5$ . Why is this significant? (2 marks) When $x=5$ , the denominator $x-5$ becomes $5-5=0$ . This makes the profit expression $\frac{2(5)}{0} = \frac{10}{0}$ , which is undefined. This is significant because it indicates a critical point where the mathematical model breaks down, possibly representing a scenario where the profit cannot be determined or a point of instability in the company's operations, such as a break-even point or a point of infinite loss/gain, depending on the context of the model.
- ii. Determine for which values of $x$ the profit is positive. (2 marks)
For the profit to be positive, $\frac{2x}{x-5} > 0$ .
This occurs when both the numerator and denominator have the same sign.
Case 1: Both are positive.
$2x > 0 \implies x > 0$
$x-5 > 0 \implies x > 5$
For both conditions to be true, $x > 5$ .
Case 2: Both are negative.
$2x < 0 \implies x < 0$
$x-5 < 0 \implies x < 5$
For both conditions to be true, $x < 0$ .
The profit is positive when $\boxed{x < 0 \text{ or } x > 5}$ .

4. a. Make the indicated variable the subject: (4 marks)
- i. Make $y$ the subject of $3x + 2y = 12$
Subtract $3x$ from both sides.
$$ 2y = 12 - 3x $$
Divide both sides by 2.
$$ y = \frac{12 - 3x}{2} $$ The subject is $\boxed{y = \frac{12 - 3x}{2}}$.
- ii. Make $r$ the subject of $A = \pi r^2 + 2\pi rh$
Rearrange the equation into the standard quadratic form $ar^2 + br + c = 0$ .
$$ \pi r^2 + 2\pi hr - A = 0 $$
Use the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a=\pi$ , $b=2\pi h$ , and $c=-A$ .
$$ r = \frac{-(2\pi h) \pm \sqrt{(2\pi h)^2 - 4(\pi)(-A)}}{2(\pi)} $$

Simplify the expression.

$$
r = \frac{-2\pi h \pm \sqrt{4\pi^2 h^2 + 4\pi A}}{2\pi}
$$
$$
r = \frac{-2\pi h \pm \sqrt{4\pi(\pi h^2 + A)}}{2\pi}
$$
$$
r = \frac{-2\pi h \pm 2\sqrt{\pi(\pi h^2 + A)}}{2\pi}
$$

Divide all terms by $2\pi$ .
$$ r = -h \pm \frac{\sqrt{\pi(\pi h^2 + A)}}{\pi} $$ The subject is $\boxed{r = -h \pm \frac{\sqrt{\pi(\pi h^2 + A)}}{\pi}}$.
4. b. Solve the following equations: (6 marks)
- i. $5x - 3 = 2x + 9$
Subtract $2x$ from both sides.
$$ 5x - 2x - 3 = 9 $$ $$ 3x - 3 = 9 $$
Add 3 to both sides.
$$ 3x = 9 + 3 $$ $$ 3x = 12 $$

Divide by 3.

$$
x = \frac{12}{3}
$$
$$
x = 4
$$
The solution is $\boxed{x = 4}$.

d. A square piece of land has an area of x2 square metres. A smaller square of area 25 square metres is removed from one corner. Write an expression for the remaining area and factorise it. (3 marks)

Step-by-step solution

Write the expression for the remaining area.

Factorise the expression using the difference of two squares formula, $A^2 - B^2 = (A - B)(A + B)$ .

Find a common denominator, which is $(x+1)(x+2)$ .

Combine the fractions and expand the numerator.

Simplify the numerator.

Factorise the numerators and denominators.

Change division to multiplication by the reciprocal of the second fraction.

Cancel common factors $(2m+3)$ and $(m-1)$ .

A fraction is undefined when its denominator is equal to zero.

Solve for $x$ .

A fraction is zero when its numerator is equal to zero, provided the denominator is not also zero for that value of $x$ .

Solve for $x$ .

Check if these values make the denominator zero.

For the profit to be positive, $\frac{2x}{x-5} > 0$ .

Subtract $3x$ from both sides.

Divide both sides by 2.

Rearrange the equation into the standard quadratic form $ar^2 + br + c = 0$ .

Use the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a=\pi$ , $b=2\pi h$ , and $c=-A$ .

Simplify the expression.

Divide all terms by $2\pi$ .

Subtract $2x$ from both sides.

Add 3 to both sides.

Divide by 3.

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More Mathematics Questions

d. A square piece of land has an area of x2 square metres. A smaller square of area 25 square metres is removed from one corner. Write an expression for the remaining area and factorise it. (3 marks)

Step-by-step solution

Write the expression for the remaining area.

Factorise the expression using the difference of two squares formula, A2−B2=(A−B)(A+B)A^2 - B^2 = (A - B)(A + B)A2−B2=(A−B)(A+B).

Find a common denominator, which is (x+1)(x+2)(x+1)(x+2)(x+1)(x+2).

Combine the fractions and expand the numerator.

Simplify the numerator.

Factorise the numerators and denominators.

Change division to multiplication by the reciprocal of the second fraction.

Cancel common factors (2m+3)(2m+3)(2m+3) and (m−1)(m-1)(m−1).

A fraction is undefined when its denominator is equal to zero.

Solve for xxx.

A fraction is zero when its numerator is equal to zero, provided the denominator is not also zero for that value of xxx.

Solve for xxx.

Check if these values make the denominator zero.

For the profit to be positive, 2xx−5>0\frac{2x}{x-5} > 0x−52x​>0.

Subtract $3x$ from both sides.

Divide both sides by 2.

Rearrange the equation into the standard quadratic form ar2+br+c=0ar^2 + br + c = 0ar2+br+c=0.

Use the quadratic formula r=−b±b2−4ac2ar = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}r=2a−b±b2−4ac​​, where a=πa=\pia=π, b=2πhb=2\pi hb=2πh, and c=−Ac=-Ac=−A.

Simplify the expression.

Divide all terms by 2π2\pi2π.

Subtract $2x$ from both sides.

Add 3 to both sides.

Divide by 3.

Still stuck on this one?

More Mathematics Questions

Factorise the expression using the difference of two squares formula, $A^2 - B^2 = (A - B)(A + B)$ .

Find a common denominator, which is $(x+1)(x+2)$ .

Cancel common factors $(2m+3)$ and $(m-1)$ .

Solve for $x$ .

A fraction is zero when its numerator is equal to zero, provided the denominator is not also zero for that value of $x$ .

Solve for $x$ .

For the profit to be positive, $\frac{2x}{x-5} > 0$ .

Rearrange the equation into the standard quadratic form $ar^2 + br + c = 0$ .

Use the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a=\pi$ , $b=2\pi h$ , and $c=-A$ .

Divide all terms by $2\pi$ .