Given that \(x_1\) and \(x_2\) are the roots of \(ax^2 + bx + c = 0\), state in terms of some or all of \(a\), \(b\), \(c\): (i) the condition that \(x_1 = x_2\),
We know that \(a \neq 0\), since otherwise the equation would only have one root.
The discriminant of the quadratic equation \(ax^2 + bx + c = 0\) is given by \(b^2 - 4ac\).
There is one repeated root (that is, \(x_1 = x_2\)) if and only if \(b^2 - 4ac = 0\), or equivalently \(b^2 = 4ac\).
(ii) the value of \(x_1 + x_2\).
Since the roots of the quadratic equation \(ax^2 + bx + c = 0\) are \(x_1\) and \(x_2\), we can rewrite the equation as \(a(x - x_1)(x - x_2) = 0\).
Expanding this expression then yields \[ax^2 - a(x_1 + x_2)x + ax_1x_2 = 0.\] Comparing these coefficients with the ones in the original equation yields \(b = -a(x_1 + x_2)\) and hence \(x_1 + x_2 = -\dfrac{b}{a}\).
- Find the values of \(m\) for which the line \(y = mx\) is a tangent to the curve \(y^2 = 3x - 1\).
The line \(y = mx\) is a tangent to the curve \(y^2 = 3x - 1\) if the line touches the curve at exactly one point.
The intersection points are found by solving the equations of the line and the curve simultaneously.
Substituting \(y=mx\) into the equation of the curve gives \(m^2 x^2 = 3x - 1\), which is equivalent to the equation \(m^2x^2 - 3x + 1 = 0\).
By part (i) above, we know that we have exactly one root if and only if \(9 = 4m^2\), so \(m = \pm \dfrac{3}{2}\).
- The line \(y = 2x\) meets the curve \(3y = x^2 - 10\) at the points \(A(x_1, y_1)\) and \(B(x_2, y_2)\).
- Obtain the quadratic equation whose roots are \(x_1\) and \(x_2\).
As in part (a), the intersection points of a line with a curve are obtained by solving their equations simultaneously.
To find their \(x\)-coordinates, we substitute \(y=3x\) into the equation of the curve to give \(6x=x^2-10\), which rearranges to \(x^2 - 6x - 10 = 0\).
- Without solving this equation, find the \(x\) co-ordinate of the midpoint of \(AB\).
We know that the \(x\) coordinates of the points \(A\) and \(B\) are \(x_1\) and \(x_2\) respectively, so the midpoint of \(AB\) has \(x\)-coordinate \(\dfrac{x_1+x_2}{2}\).
By part (ii) above, we know that \(x_1 + x_2 = 6\), since they are the roots of the equation \(x^2 -6x - 10 = 0\). Thus the midpoint of \(AB\) has \(x\)-coordinate \(\dfrac{6}{2}=3\).
