Find the values of k for which the line y=1-2kx does not meet the curve y=9x²-(3k+1)x+5

Let's equate the two given functions and attempt to solve for x:

y = 1 -2kx = y = 9x^2 -(3k+1)x + 5

Eliminating y, 1 -2kx = 9x^2 -(3k+1)x + 5

Rearranging terms in descending order by powers of x:

0 = 9x^2 - (3k+1)x + 2kx + 5 - 1 , or

0 = 9x^2 - kx - x + 4

This is a quadratic equation with coefficients a = 9, b = -(k+1) and c = 4.

For certain k, not yet known, solutions exist. Solutions here implies points at which the two curves intersect.

k+1 plus or minus sqrt( [-(k+1)]^2 - 4(9)(4) )

x =

2(9)

The discriminant is k^2 + 2k + 1 - 144, or k^2 + 2k - 143.

If the discriminant is > 0, there are two real, unequal roots. We don't want this, since we're interested in finding k value(s) for which there's no solution.

If the discr. is = 0, there are two real, equal roots. Again, we don't want this.

If the discr. is < 0, there are no real roots. This is the case that interests us.

So our final task is to determine the k values for which the discr. is < 0:

Determine the k value(s) for which the discriminant, k^2 + 2k - 143, is 0.

This k^2 + 2k - 143 factors as follows: (k-11)(k+13), and when set = to 0, results in k: {-13,11}.

Set up intervals on the number line: (-infinity, - 13), (-13, 11) and (11, infinity).

Choosing a test number from each interval, determine the interval or intervals on which the discriminant is negative:

Case 1: k = -15; the discriminant (k^2 + 2k - 143) is (-15)^2 + 2(-15) - 143 = +52. Reject this interval

Case 2: k = 0; the discriminant is then 0 + 0 - 143 (negative); thus, the discriminant is negative on the interval (-13,11).

Case 3: k = 20; the discriminant is positive. Reject this interval.

Summary: The curves do not intersect on the interval (-13,11).

The discriminant of the function must be negative, for no solution to exist.

x= 149. 24+7=31. 180-31=149

your welcome

the mistake in the pattern is that :

b. the decimal point was moved one too many place values each time

the correct pattern could be made by :

b. moving the decimal point to the right the same numbers of place values as the exponents