f.4 mathz,,,find the two possible values of the constanr k &

a.

y=(2k+1)x^2-4(k-2)x+16 touches the x-axis at only one point.
=> y = 0
=. 0 = (2k+1)x^2-4(k-2)x+16
一元二次方程式 (-b + - ((b^2 -4ac))^0.5) / 2a 中的 b^2 -4ac = 0 因為X只得一個值。

(-4(k-2))^2 - 4 (16)(2k+1) = 0
16(k-2)^2 - 4(16)(2k+1) = 0
(k-2)^2 - 4(2k+1) = 0
k^2-12k = 0
k = 0 or k = 12

b.
If k is positive => 12 then

0 = (2*12+1)x^2-4(12-2)x+16
0 = 25x^2-40x+16
0 = (5x-4)(5x-4)
x = 4/5