Simultaneous Linear and Quadratic Equations - SS2 Mathematics Past Questions and Answers - page 1
Solve the equation \(x^{2} + 5x + 4 = 0\) using the completing the square method
x = 2 or 1
x = 1 or -4
x = -4 or -1
x = 2 or -2
\[x^{2} + 5x + 4 = 0\]
\[x^{2} + 5x = - 4\]
\[x^{2} + 5x + \left( \frac{5}{2} \right)^{2} = \left( \frac{5}{2} \right)^{2} - 4\]
\[x^{2} + 5x + \frac{25}{4} = \frac{25}{4} - 4\]
\[{(x + \frac{5}{2})}^{2} = \frac{9}{4}\]
\[x + \frac{5}{2} = \pm \sqrt{\frac{9}{4}} = \pm \frac{3}{2}\]
\[x + \frac{5}{2} = \pm \frac{3}{2}\]
\[x = \frac{3}{2} - \frac{5}{2}\ or - \frac{3}{2} - \frac{5}{2}\]
\[x = \frac{3 - 5}{2}\ or\frac{- 3 - 5}{2}\]
\[x = \frac{- 2}{2}\ or\frac{- 8}{2}\]
\(x = - 1\ or - 4\)
Solve the simultaneous equations \(x + y = 8\) and \(x^{2} - y^{2} = 16\).
(0; - 3) or ( - 7;5)
( - 4;3) or (2;5)
(5;3)
(3;3) or (5;5)
\(x + y = 8\) (1)
\(x^{2} - y^{2} = 16\) (2)
From (1), transpose \(y\)
\[x + y = 8\]
\[y = 8 - x\]
Substitute the value of \(y\) in (2)
\[x^{2} - {(8 - x)}^{2} = 16\]
\[16x - 64 = 16\]
\[16x = 16 + 64\]
\[16x = 80\]
\[16x - 80 = 0\]
\[16(x - 5) = 0\]
\[\therefore x = 5\]
Substitute the values of \(x\) in (1)
\[x + y = 8\]
\[5 + y = 8\]
\[y = 8 - 5 = 3\]
The solution to the simultaneous equations \(2x + y = 5\) and \(x^{2} + y^{2} = 25\) is the ordered pairs: \((5;3)\)
Solve the equation \({2x}^{2} + 9x = 5\) using the completing the square method
\[{2x}^{2} + 9x = 5\]
\[x^{2} + \frac{9}{2}x = \frac{5}{2}\]
\[x^{2} + \frac{9}{2}x + {(\frac{9}{4})}^{2} = \frac{5}{2} + {(\frac{9}{4})}^{2}\]
\[x^{2} + \frac{9}{2}x + \frac{81}{16} = \frac{5}{2} + \frac{81}{16}\]
\[{(x + \frac{9}{4})}^{2} = \frac{5}{2} + \frac{81}{16}\]
\[{(x + \frac{9}{4})}^{2} = \frac{40 + 81}{16}\]
\[{(x + \frac{9}{4})}^{2} = \frac{121}{16}\]
\[x + \frac{9}{4} = \pm \sqrt{\frac{121}{16}} = \pm \frac{11}{4}\]
\[x = - \frac{9}{4} \pm \frac{11}{4}\]
\[x = - \frac{9}{4} + \frac{11}{4}\ \ \ \ \ or\ \ \ \ \ - \frac{9}{4} - \frac{11}{4}\]
\(x = \frac{11 - 9}{4}\ \ \ \ or\ \frac{- 9 - 11}{4}\)
\(x = \frac{1}{2}\) or \(- 5\)
Solve the simultaneous equations \(3x + y = 10\) and \({2x}^{2} + y^{2} = 19\).
\(3x + y = 10\) (1)
\({2x}^{2} + y^{2} = 19\) (2)
From (1), transpose \(y\),
\[3x + y = 10\]
\[y = 10 - 3x\]
In (2), substitute the value of \(y\),
\[{2x}^{2} + {(10 - 3x)}^{2} = 19\]
\[{2x}^{2} + {9x}^{2} - 60x + 100 = 19\]
\[{11x}^{2} - 60x + 100 - 19 = 0\]
\[{11x}^{2} - 60x + 81 = 0\]
\[\therefore x = 3\ or\ \frac{27}{11}\]
Substitute the values of \(x\) in (1)
\[3x + y = 10\]
For \(x = 3\)
\[9 + y = 10\]
\[y = 10 - 9 = 1\]
For \(x = \frac{27}{11} = 2.45\)
\[\frac{27}{11} + y = 10\]
\[y = 10 - \frac{27}{11}\]
\[y = 7.55\]
The solution to the simultaneous equations \(3x + y = 10\) and \({2x}^{2} + y^{2} = 19\) are the ordered pairs: \((3;1)or\ (2.45;7.55)\)