(a) The roots of the equation \(x^{2} + mx + 11 =... | WAEC SSCE - Further Mathematics - 2007

Question 1 Report

(a) The roots of the equation \(x^{2} + mx + 11 = 0\) are \(\alpha\) and \(\beta\), where m is a constant. If \(\alpha^{2} + \beta^{2} = 27\), find the values of m.

(b) The line \(2x + 3y = 1\) intersects the circle \(2x^{2} + 2y^{2} + 4x + 9y - 9 = 0\) at points P and Q where Q lies in the fourth quadrant. Find the coordinates of P and Q.

Answer

(a) For \(x^2 + mx + 11 = 0\) with roots \(\alpha,\beta\): \(\alpha + \beta = -m\) and \(\alpha\beta = 11\).

\[\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = m^2 - 22\]

Setting this equal to 27:

\[m^2 - 22 = 27 \ \Rightarrow\ m^2 = 49 \ \Rightarrow\ m = \pm 7\]

(b) From the line \(2x + 3y = 1\), \(x = \dfrac{1 - 3y}{2}\). Substitute into the circle \(2x^2 + 2y^2 + 4x + 9y - 9 = 0\) (dividing by 2 first: \(x^2 + y^2 + 2x + 4.5y - 4.5 = 0\)):

\[\frac{(1 - 3y)^2}{4} + y^2 + (1 - 3y) + 4.5y - 4.5 = 0\]

Multiply through by 4 and simplify:

\[13y^2 - 13 = 0 \ \Rightarrow\ y^2 = 1 \ \Rightarrow\ y = \pm 1\]

For \(y = 1\): \(x = \dfrac{1 - 3}{2} = -1\), giving \((-1, 1)\).

For \(y = -1\): \(x = \dfrac{1 + 3}{2} = 2\), giving \((2, -1)\), which lies in the fourth quadrant.

Therefore \(P = (-1, 1)\) and \(Q = (2, -1)\).

Read lesson note on Probability (WAEC)

Read lesson note on Polynomial Functions (WAEC)

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