1) find the values of k which the line y=kx-3 intersects the curve y=x2-9x at two distinct points

Question

i dont know how to do these

and help me do this one too

2) find the values of k which the line y=2x+k meets the curve y=1+2kx-x² at two distinct points

3) A curve has equation y=2xy+5 and a line has equation 2x+5y=1

The curve and the line intersect at the pointa A and B. Find the coordinates of midpoint of the line AB

4) a) Express 9x² -15x in the form (3x-a)²-b

b) Find the set of values of x that satisfy the inequality 9x²-15x<6

Answer 1

1) x^2 - 9x = kx - 3

x^2 - 9x - kx + 3 = 0

x^2 - (9+k)x + 3 = 0

A=1 B = -(k+9) and C=3

two distinct solutions required means the discriminant must be positive.

B^2 - 4ac > 0

(k+9)^2 - 4(1)(3) > 0

(k+9)^2 - 12 ,> 0

k^2 + 18k + 81 - 12 > 0

k^2 + 18k + 69 > 0

k = [-18 +or- sqrt( 18^2 - 4(1)(69)]/2

= [-18 +or- sqrt( 324 - 276]]/2

= [-18 +or- sqrt( 48)]/2

= [-18 +or- 4*Sqrt(3)]/2

= -9 +or- 2*Sqrt(3)

which are approximately x=-5.535898384862245412945107319883...

and x=-12.46410161537754587054892683012...

for x=-10, (-10)^2 +18(-10) + 69 = 100 - 180 + 69 = -80 + 69 = -11 <0, so the function is negative.

so on the interval (-infinity , -9 - 2 * sqrt(3) ) U ( -9 + 2 * sqrt(3) the line and parabola will meet at two distinct points.

======================================================

#2)

2x + k = 1 + 2kx - x^2

x^2 + 2x - 2kx - 1 = 0

x^2 -x( 2 + 2k) - 1 = 0

A=1 B= -(2+2k) and C = -1

again, for there to be two distinct solutions, the disrimminant must be positive.

(2+2k)^2 - 4(1)(-1) > 0

4 + 8k + 4k^2 + 4 > 0

4k^2 + 8k + 8 > 0

k^2 + 2k + 2 > 0

k = (-2 +or- sqrt( 4 - 8))/2 = (-2 +or- sqrt(4 - 8))/2 = (-2 +or- sqrt(-4))/2 = (-2 +or- 2*i)/2 = -1 +or- i

so there are no such x where this function is zero.

the function is positive everywhere

so there are always two unique solutions for any value of k

========================================================================

#3)

2x = 1 - 5y

substitutes into first equation for 2x

y = (1 - 5y)y + 5

y = y - 5y^2 + 5

0 = 5 - 5y^2

0 = 1 - y^2

y^2 = 1

y = +or- 1

Substituting into the equation above in bold, the two solutions are (-2,1) and ( 3,-1)

which are A and B respectively. the midpoint is ( 1/2, 0)

==============================================

#4)

9x^2 - 15x = 9(x^2 - (5/3)x ) = 9( x^2 - (5/3)x + 25/36) - 25/4

= 9( x - 5/6)^2 - 25/4

= (3*3)(x-5/6)^2 - 25/4

= (3x - 5/2)^2 - 25/4

check:

(3x - 5/2)(3x - 5/2) - 25/4 = 9x^2 -30/2x + 25/4 - 25/4 = 9x^2 - 15x

9x^2 - 15x - 6 < 0

3x^2 - 5x - 2 < 0

the solutions are:

(5 +or - sqrt( 25 + 24))/6

(5 +or sqrt(49))/ 6 =

(5 +or- 7) / 6 =

{2, -1/3}

the intervals of interest are (-infinity,-1/3) U (-1/3, 2) U ( 2, infinity)

the function is negative, hence the solution, on (-1/3,2)