Question

for what value(s) of k will the function h(x) = 4x^2 -kx +25 have only one zero?

Answer 1

h(x) will have one zero only when the quadratic can be factored into a perfect square.
Since you have the term 4x^2= (2x)^2, and u have 25=5^2
so h(x) will have one zero when h(x)= (2x-5)^2 or h(x)=(2x+5)^2
i.e. h(x)= 4x^2-20x+25 or h(x)= 4x^2+20x+25

so for k=+20 and k=-20

Answer 2

i got +/- 20 too but doesnt that make it have 2 zeros?

Answer 3

no the two are separate cases, separate functions. k=20 gives an entirely separate h(x) function from k=-20

Answer 4

if you sub it back in you get x =5/2 and x = -5/2

Answer 5

like if you factor it then you get those two answers right? and doesnt that make it 2 zeros? so confused

Answer 6

see what happens is for k=0,k=1,k=2,k=3 you get a different function every time since the expression of h(x) changes with each value of k. What the question asks you to find out is-" In how many such cases will h(x) have just one zero?"
When k=-20, h(x)= (2x+5)^2, right? so in this case h(x)=0 only at one point, x=-5/2

similarly when k=20, h(x)=0 only at one point, x=5/2..
the two are separate functions entirely

Answer 7

Ohhh right, okay thank you so much! I understand it now