The quadratic equation (k+1)x^2 + (5k - 3)x + 3k = 0 has equal roots. Find the possible values of k

We know the discriminant (b^2 - 4ac) must be equal to zero for an equation to have equal roots (think about the fact that the square root of this is taken in the quadratic equation). So we can form the equation (5k-3)^2 - 4(k+1)(3k) = 0Simplifying this to 13k^2 - 42k + 9 = 0 and factorising to (13k - 3)(k - 3) = 0 (easily done by spotting that 13 is prime), we can see that k = 3 or k = 3/13

MI

Related Maths A Level answers

All answers ▸

find the coordinates of the turning points of the curve y = 2x^4-4x^3+3, and determine the nature of these points


Find the derivative of e^3x


A curve C has equation y = x^2 − 2x − 24 x^(1/2), x > 0 (a) Find (i) dy/d x (ii) d^2y/dx^2 (b) Verify that C has a stationary point when x = 4 (c) Determine the nature of this stationary point, giving a reason for your answer.


A curve has equation y=x^2 + (3k - 4)x + 13 and a line has equation y = 2x + k, where k is constant. Show that the x-coordinate of any point of intersection of the line and curve satisfies the equation: x^2 + 3(k - 2)x + 13 - k = 0