values for which quadratic curve lies below x axis

Question

A quadratic equation $y=(k+1)x^2-3x+(k+1)$ we need to the find the set of values of $k$ for which the curve $y$ lies below the $x-$ axis.

I used the quadratic formula and equate it to $0$

$ 3\pm \frac{\sqrt{ 9-4(k+1)^2}}{2(k+1)}=0 $

PS assist, how to solve to get further.

Thanks, metric

Answer 1

Let's calculate the discriminant. $\Delta = 9-4(k+1)^{2}$. We must have $\Delta<0$ and the coefficient in $x^{2}$ strictly negative. so $\frac{9}{4}<(k+1)^{2}$. So $k+1<-\frac{3}{2}$ or $k+1>\frac{3}{2}$ and we must have $k+1<0$.So $k<-\frac{5}{2}$.

Answer 2

thanks for the hint, I understand this is how it should be

curve below x-axis (should be inverted parabola) and no real solution

Discriminant: $~~~9-4(k+1)^2<0~~~$ or $~~~4k^2+8k-5 >0$

solving above inequality, $k<\frac{-5}{2} , k>\frac{1}{2}$

hope this is correct.

Thanks.

Answer 3

From your description, it is hard to determine which case you are looking for.

For case 2, we want for ALL x such that the corresponding y is below the x-axis. In this case, we take ⊿ < 0 together with k + 1 < 0.

For case 1, we need to find the roots ($\alpha, \beta$ with $\alpha < \beta$, say) first. The discussion must then be separated into case 1a and case 1b depending on the value of k. For case 1a, the successful candidates are those x’s such that $x < \alpha$ OR $x > \beta$. For case 1b, …..

So, case 1 or 2?

PS. Your question did not say any restriction on k too. What would happen if $k = –1$?