if the equation x^3-hx^2+kx-9=0 has only one real root which has a value of 1, find the range of values of k.

Question

empty · Answer

what can h be, just any real number?

anonymous · Answer

not mention

empty · Answer

Well here's how I solved it, just start with the factored form, since we know one of the roots is 1, but we also have two other roots too: $$(x-a)(x-b)(x-1)=x^3-hx^2+kx-9$$ Now when we multiply it out on the left side we get: $$x^3-(a+b+1)x^2+(ab+a+b)x-ab=x^3-hx^2+kx-9$$ now we can just set the coeffients equal to each other! $$h=a+b+1 \ ab+a+b=k \ ab=9$$ Now look at the equation for k in the middle, we can subtract 1 from the first equation to get a+b nd we have ab from the second equation, so we can just plug them in like that $$9+(h-1)=k $$

anonymous · Answer

the answer is 3 < k <15

rational · Answer

$$x^3-hx^2+kx-9=0$$

since $1$ is a root, it must satisfy above eqn 
$$1-h+k-9 = 0\h =k-8$$
the eqn becomes
$$x^3-(k-8)x^2 + kx -9$$

The cubic will have only one root after and before the graph touches x axis. That is when the other root $r$ has a multiplicity of $2$ : 
$$x^3-(k-8)x^2 + kx -9 = (x-1)(x-r)^2$$
Comparing constant terms we get $r = \pm 3$

comparing $x$ coefficients we get $k = 2r+r^2$

use $r$ values for getting the bounds of $k$