Ok i got number 3, i realised that delta = 0 when there is a TP on x-axis.
So k = +-2 root 6.
Hi! Thanks for helping me!
Show that the equation x^3 - x^2 +x - 21 = 0 has only one solution.
Without dividing find the remainder when x^3 - 4x + 12x - 2 is divided by x+1.
and one more
For what values of k does the graph of y = 2x^2 - kx + 3 have a TP on the x-axis?
Thanks for your great help!
You can easily show it by demonstrating that this function is increasing only, and hence intersects the x-axis exactly one. Thus, it has only one real solution.Originally Posted by classicstrings
But you need to have the knowledge of calculus, I do not know if you studied it. Did you?
Hello,Originally Posted by classicstrings
your equation consists of a term in x T(x) which is equal to zero. If there is a solution you can split this term into a linear factor (x-s) (s stands for solution) and the remaining term R(x). So you get . When you multiply the RHS of this equation you can see, that s must be a divisor of the constant summand of your equation. All integer divisors of 21 are: +-1, +-3, +-7, +-21.
Now do a little bit trial and error and you'll find pretty soon, that 3 must be a solution of your equation. That means you can divide the LHS of your equation by (x-3):
Because for all there are no other possible solutions.