Exam Question

Curve: y=x^2+7 , Line: y=k(3x+1)

• a). Show that the x-coordinates of any points of intersection of C and L satisfy: x^2-3kx+7-k=0
• b). C and L intersect at 2 distinct points. Show that: 9k^2+4k-28>0
• c). Solve the inequality: 9k^2+4k-28>0

Curve: y=x^2+7 , Line: y=k(3x+1)

• a). Here we read "points of intersection of C and L" and notice that C and L both equal y, so we set them equal to each other: x^2+7=k(3x+1). This can also be written as x^2+7=3xk+k. Now we rearrange to make it equal 0 giving us x^2-3kx-k+7=0, and the question is complete.
• b). Here we are told C and L intersect at 2 distinct points, so we know the discriminant of the single equation x^2-3kx-k+7=0 must be greater than 0 (b^2-4ac>0). Now we sub in the values so: (-3k)^2-4x1x(-k+7)>0 , and this expands to 9k^2+4k-28>0, and the question is complete.
• c). Here we just solve by either factorising, completing the square or using the quadratic formula. Here I will factorise, so 9k^2+4k-28=0 is equal to (9k-14)(k+2) , which has the solutions k=14/9 or k=-2. Now we express this as an inequality, so -2<k<14/9. The question is complete.