(3 math quest.)The graph of function f(x)=x² - kx+k+8 touches the x-axis at one point. What is the value of k?

Question

Yeah, so I have 2 other questions to answer as well.

1) Show that, if k is any real number, kx² + (3k+2)x +(2k+3) = 0

2) For what values of m does the equation have two equal roots?
x² - 15 = 2m(x-4)

Anyway, any help on how to do this work would be great, and very appreciated

Answer 1

Both f(x) and f'(x) have to be xero. Write these 2 eqns out and solve simultaneously for x & k I got 2 values for each.....

1) If k is real and x can have any value, this equality is easy to disprove with a single example.

2) 2 equal roots means the same as touching the x axis at one point, as in your initial question. Work it out the same way....

Answer 2

You must use the binomial theorem for these questions. The first question can be answered by setting b^2-4ac=0, so (1-k)^2-4k(2k+3)=0,solve for k. For the second question, b^2-4ac must be greater than or equal to zero. For the third question, I don't get it, since if both roots are equal, then they are the same root, aren't they?

Answer 3

If the graph of the functionality of f (x) touches the x-axis only as quickly as, then the functionality might desire to purely have one 0 (do you notice why?) subsequently, the discriminate of the quadratic functionality might desire to equivalent 0. So: ?(ok² - 4(ok + 8)) = 0 ok² - 4k - 32 = 0 (ok - 8)(ok + 4) = 0 ok = 8, ok = -4.