(a) Solve the equation \(8x^2 – 11x – 11 = 0\).
Show all your working and give your answers correct to 2 decimal places.
(b) y varies directly as the square root of x.
y = 18 when x = 9.
Find y when x = 484.
(c) Sara spends $x on pens which cost $2.50 each.
She also spends $(x – 14.50) on pencils which cost $0.50 each.
The total of the number of pens and the number of pencils is 19.
Write down and solve an equation in x.
▶️ Answer/Explanation
(a) Ans: \(x = -0.67\) and \(x = 2.05\)
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\), where \(a = 8\), \(b = -11\), and \(c = -11\). The discriminant is \(\sqrt{(-11)^2 – 4(8)(-11)} = \sqrt{121 + 352} = \sqrt{473}\). Thus, the solutions are \(x = \frac{11 \pm \sqrt{473}}{16}\), giving \(x \approx -0.67\) and \(x \approx 2.05\).
(b) Ans: \(y = 132\)
Given \(y = k\sqrt{x}\), substitute \(y = 18\) and \(x = 9\) to find \(k = 6\). For \(x = 484\), \(y = 6 \times \sqrt{484} = 6 \times 22 = 132\).
(c) Ans: \(x = 20\)
Number of pens = \(\frac{x}{2.5}\), pencils = \(\frac{x – 14.5}{0.5}\). Total is \(\frac{x}{2.5} + \frac{x – 14.5}{0.5} = 19\). Multiply through by 2.5 to solve: \(x + 5x – 72.5 = 47.5\), simplifying to \(6x = 120\) and \(x = 20\).
Bernie buys x packets of seeds and y plants for his garden.
He wants to buy more packets of seeds than plants.
The inequality \(x > y\) shows this information.
He also wants to buy
• less than 10 packets of seeds
• at least 2 plants.
(a) Write down two more inequalities in x or y to show this information.
(b) Each packet of seeds costs $1 and each plant costs $3.
The maximum amount Bernie can spend is $21.
Write down another inequality in x and y to show this information.
(c) The line \(x = y\) is drawn on the grid.
Draw three more lines to show your inequalities and shade the unwanted regions.
(d) Bernie buys 8 packets of seeds.
(i) Find the maximum number of plants he can buy.
(ii) Find the total cost of these packets of seeds and plants.
▶️ Answer/Explanation
(a) Ans: \(x < 10\) and \(y \geq 2\)
Bernie wants fewer than 10 seed packets (\(x < 10\)) and at least 2 plants (\(y \geq 2\)).
(b) Ans: \(x + 3y \leq 21\)
Total cost constraint: \(1x + 3y \leq 21\) (seeds cost $1 each, plants cost $3 each).
(c)
Draw lines for \(x = 10\) (vertical), \(y = 2\) (horizontal), and \(x + 3y = 21\) (from (0,7) to (21,0)). Shade regions violating the inequalities.
(d)(i) Ans: 4
With \(x = 8\), from \(8 + 3y \leq 21\), \(y \leq 4.33\). Maximum plants: 4.
(d)(ii) Ans: 20
Total cost: \(8 \times 1 + 4 \times 3 = 8 + 12 = \$20\).