• It is given that the simultaneous equations $$\left\{ \begin{array}{1}y = {x^2} - kx + 34\\y = 6x - 3k\end{array} \right.$$ have only one set of real solution and k is a positive constant.
Solve the simultaneous equations.
• $$\left\{ \begin{array}{l}y = {x^2} - kx + 34\;.....................\;(1)\\y = 6x - 3k\;............................\;(2)\end{array} \right.$$
Substitute (2) into (1),
$$\begin{array}{1}6x - 3k = {x^2} - kx + 34\\{x^2} - (k + 6)x + 3k + 34 = 0\;........\;(3)\end{array}$$
∵ The simultaneous equations have only one set of real solution.
i.e. Equation (3) has only one double real root.
∴ $$\Delta = 0$$
i.e. $$\begin{array}{1}{[ - (k + 6)]^2} - 4(1)(3k + 34) = 0\\{k^2} + 12k + 36 - 12k - 136 = 0\\{k^2} = 100\end{array}$$
$$k = 10$$ or $$k = - 10$$ (rejected)
Substitute k = 10 into (3),
$$\begin{array}{1}{x^2} - (10 + 6)x + 3(10) + 34 = 0\\{x^2} - 16x + 64 = 0\\{(x - 8)^2} = 0\\x = 8\end{array}$$
Substitute k = 10 and x = 8 into (2),
$$\begin{array}{1}y = 6(8) - 3(10)\\ = 18\end{array}$$
∴ The solution is $$x = 8$$, $$y = 18$$.

• (a) Solve the simultaneous equations $$\left\{ \begin{array}{l}y = {x^2} + 4x + 2\\y - 2x - 2 = 0\end{array} \right.$$.
(b) Using the result of (a), solve the simultaneous equations $$\left\{ \begin{array}{l}a + b = {(a - b)^2} + 4(a - b) + 2\\3b - a - 2 = 0\end{array} \right.$$.
• (a) $$\left\{ \begin{array}{l}y = {x^2} + 4x + 2\;.........\;(1)\\y - 2x - 2 = 0\;...........\;(2)\end{array} \right.$$
Substitute (1) into (2),
$$\begin{array}{1}{x^2} + 4x + 2 - 2x - 2 = 0\\{x^2} + 2x = 0\\x(x + 2) = {0^{}}\end{array}$$
$$x = 0$$ or $$x = - 2$$
Substitute x = 0 into (1), $$\begin{array}{1}y = {(0)^2} + 4(0) + 2\\ = 2\end{array}$$
Substitute x = -2 into (1), $$\begin{array}{1}y = {( - 2)^2} + 4( - 2) + 2\\ = - 2\end{array}$$
∴ The solutions are $$x = 0$$, $$y = 2$$ and$$x = - 2$$, $$y = - 2$$.

(b) Rewrite $$\left\{ \begin{array}{l}a + b = {(a - b)^2} + 4(a - b) + 2\\3b - a - 2 = 0\end{array} \right.$$ as $$\left\{ \begin{array}{l}a + b = {(a - b)^2} + 4(a - b) + 2\\(a + b) - 2(a - b) - 2 = 0\end{array} \right.$$.
Let $$x = a - b$$ and$$y = a + b$$, then the simultaneous equations become $$\left\{ \begin{array}{l}y = {x^2} + 4x + 2\\y - 2x - 2 = 0\end{array} \right.$$.
Using the result of (a), we have
$$\left\{ \begin{array}{l}x = 0\\y = 2\end{array} \right.$$ or $$\left\{ \begin{array}{l}x = - 2\\y = - 2\end{array} \right.$$
i.e. $$\left\{ \begin{array}{l}a - b = 0\;........\;(3)\\a + b = 2\;........\;(4)\end{array} \right.$$ or $$\left\{ \begin{array}{l}a - b = - 2\;........\;(5)\\a + b = - 2\;........\;(6)\end{array} \right.$$
(3) + (4), $$\begin{array}{1}2a = 2\\a = 1\end{array}$$ (5) + (6), $$\begin{array}{1}2a = - {\rm{ }}4\\a = - 2\end{array}$$
(4) - (3), $$\begin{array}{1}2b = 2\\b = 1\end{array}$$ (6) - (5), $$\begin{array}{1}2b = 0\\b = 0\end{array}$$
∴ The solutions are $$a = 1$$, $$b = 1$$ and $$a = - 2$$, $$b = 0$$.

• The figure shows the graphs of $$y = 2{m^2}$$ and $$y = {x^2} - 5mx + 6{m^2}$$, where m is a positive constant.
(a) Express the coordinates of C and D in terms of m.
(b) Express the coordinates of A and B in terms of m.
(c) If the area of trapezium ABCD is 32, find the value of m
• (a) For $$y = {x^2} - 5mx + 6{m^2}$$, when $$y = 0$$,
$$\begin{array}{1}{x^2} - 5mx + 6{m^2} = 0\\(x - 2m)(x - 3m) = 0\end{array}$$
$$x = 2m$$ or $$x = 3m$$
∴ The coordinates of C and D are $$(2m{\rm{ }},{\rm{ }}0)$$ and $$(3m{\rm{ }},{\rm{ }}0)$$ respectively.

(b) $$\left\{ \begin{array}{l}y = {x^2} - 5mx + 6{m^2}\;........\;(1)\\y = 2{m^2}\;..........................\;(2)\end{array} \right.$$
Substitute (2) into (1),
$$\begin{array}{1}2{m^2} = {x^2} - 5mx + 6{m^2}\\{x^2} - 5mx + 4{m^2} = 0\\(x - m)(x - 4m) = {0^{}}\end{array}$$
$$x = m$$ or $$x = 4m$$
∴ The coordinates of A and B are $$(m{\rm{ }},{\rm{ }}2{m^2})$$ and $$(4m{\rm{ }},{\rm{ }}2{m^2})$$ respectively.

(c) Area of trapezium ABCD = 32
$$\begin{array}{1}\frac{1}{2} \times [(4m - m) + (3m - 2m)] \times 2{m^2} = 32\\4{m^3} = 32\\{m^3} = 8\\{m^3} = {2^3}\\m = \underline{\underline 2} \end{array}$$