WAEC Mathematics 2020 Theory — Question 13

Model answer

(a) Steps/procedures for construction: (i) Draw a straight line and mark point A on the line. On the line, measure |AB|=7.5cm from A. (ii) Construct angle ABC=120 deg on B, i.e. angle ABC=120 deg. (iii) Mark off point C with |AC|=13.5cm. (iv) Bisect line AB to obtain locus l1. (v) Bisect line BC to obtain locus l2. Produce l2 to meet l1 at point M. (v) At the point of intersection of l1 and l2, place a compass with suitable radius and draw a circle which passes through the vertices of triangle ABC. (b) Given x=(1-sqrt2)/2. The expression 2x^2-2x = 2x(x-1). Substituting the value of x into the expression: = 2 x [(1-sqrt2)/2] x [(1-sqrt2)/2 - 1] = (1-sqrt2) x [(1-sqrt2-2)/2] = (1-sqrt2) x [(-1-sqrt2)/2] = [1(-1-sqrt2) - sqrt2(-1-sqrt2)]/2 = [-1-sqrt2+sqrt2+2]/2 = 1/2. For the equation 4x^2-4*sqrt3*x+3=0: Dividing through by 4: x^2 - sqrt3*x + 3/4 = 0, i.e. x^2-sqrt3*x = -3/4. Adding the square of half the coefficient of x to both sides: x^2-sqrt3*x+(sqrt3/2)^2 = -3/4+(sqrt3/2)^2 = -3/4+3/4 = 0. So (x-sqrt3/2)^2=0, giving x=sqrt3/2 (a repeated root, occurring twice).