# Given: f(x) = x + x(k - 1) + k, find the range of values of k so that f(x)0 for all real values...

## Question:

Given: {eq}f(x) = x + x(k - 1) + k, {/eq} find the range of values of k so that f(x)>0 for all real values of x.

The equation {eq}ax^2+bx+c=0 {/eq} has solutions {eq}x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a} {/eq}. If {eq}b^2-4ac<0 {/eq}, then there are only complex solutions, if {eq}b^2-4ac=0 {/eq} then there is a single real solution and if {eq}b^2-4ac>0 {/eq} there are a pair of real solutions.

Let {eq}f(x)=x+x(k-1)+k=kx+k {/eq}. This is linear and unless {eq}k=0 {/eq} it will have negative values. If {eq}k=0 {/eq}, then {eq}f(x)=0 {/eq} which is not greater than 0.

I suspect this was supposed to be {eq}f(x)=x^2+x(k-1)+k {/eq} which will be a parabola opening up and it will always be positive when it has no real solutions which will happen when

{eq}(k-1)^2-4(1)(k)<0\\ k^2-6k+1<0 {/eq}

The roots of this are {eq}\frac{6\pm \sqrt{36-4(1)}}{2(1)}=\frac{6\pm \sqrt{32}}{2}=\frac{6\pm 4\sqrt{2}}{2}=3\pm 2\sqrt{2} {/eq}

So {eq}k^2-6k+1<0 {/eq} when {eq}3-2\sqrt{2}< k<3+2\sqrt{2} {/eq} 