More about equations Examples

ExamplesIt 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. 
Solutions\(\left\{ \begin{array}{l}y = {x^2}  kx + 34\;.....................\;(1)\\y = 6x  3k\;............................\;(2)\end{array} \right.\)
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 \(x = 8\), \(y = 18\).

Examples(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 \(\left\{ \begin{array}{l}a + b = {(a  b)^2} + 4(a  b) + 2\\3b  a  2 = 0\end{array} \right.\). 
Solutions(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 \(a =  2\), \(b = 0\).

ExamplesThe figure shows the graphs of \(y = 2{m^2}\) and \(y = {x^2}  5mx + 6{m^2}\), where m is a positive constant.
[GraphMissing Latex_More about equations Q29]
(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 
Solutions(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
\(\begin{array}{1}\frac{1}{2} \times [(4m  m) + (3m  2m)] \times 2{m^2} = 32\\4{m^3} = 32\\{m^3} = 8\\{m^3} = {2^3}\\m = \underline{\underline 2} \end{array}\)