(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.
Explanation
(a) (x^{2} + mx + 11 = 0)
Roots are (alpha) and (beta).
(therefore alpha + beta = frac{-b}{a} = – m & alpha beta = frac{c}{a} = 11)
(alpha^{2} + beta^{2} = (alpha + beta)^{2} – 2alpha beta)
(27 = (-m)^{2} – 2(11))
(m^{2} = 22 + 27 = 49)
(m = pm 7)
(b) Equation of line : (2x + 3y = 1 i.e. y = frac{1 – 2x}{3})
Substitute for y in the equation of the circle:
(2x^{2} + 2(frac{1 – 2x}{3})^{2} + 4x + 9(frac{1 – 2x}{3}) – 9 = 0)
(2x^{2} + 2(frac{1 – 4x + 4x^{2}}{9} + 4x + 3(1 – 2x) – 9 = 0)
(18x^{2} + 2 – 8x + 8x^{2} + 36x + 27 – 54x – 81 = 0)
(26x^{2} – 26x – 52 = 0)
(x^{2} – x – 2 = 0)
((x – 2)(x + 1) = 0)
(x = text{2 or -1})
Substitute for x in (y = frac{1 – 2x}{3}); when x = 2,
(y = frac{1 – 2(2)}{3} = frac{-3}{3} = -1)
Coordinate = ((2, -1)).
When x = -1,
(y = frac{1 – 2(-1)}{3} = frac{3}{3} = 1)
Coordinate = ((-1, 1))
Since Q lies in the fourth quadrant, we have P(-1, 1) and Q(2, -1).