# Constraints on the roots to a quadratic equation

If both roots of the quadratic equation

$$2x^2 +kx -(k+1)=0$$

are greater than $$1$$, then $$k$$ lies in what interval?

I tried to solve this using using different graphical and algebraic method but i seem to miss a crucial insight which is stopping me from getting the answer. Please help here.

• Use the formula two find the two roots in terms of k. Then, enforce each to be greater than zero and you will get the interval for $k$. – Dunkel Jun 15 at 18:26
• Why is the point $(1,1)$ on every parabola as $k$ varies over the reals? – AmateurMathPirate Jun 15 at 21:26

## 6 Answers

Let $$x_1,x_2$$ be the two roots of your equation, so $$x_1 \geq 1$$, $$x_2 \geq 1$$.

In general, given a quadratic equation $$ax^2+bx+c=0$$ you have that $$\frac{c}{a}$$ is the product of the two roots and that $$-\frac{b}{a}$$ is the sum of the two roots.

In our case $$x_1+x_2=-\frac{k}{2}$$ and $$x_1 x_2=-\frac{k+1}{2}$$ and $$x_1+x_2 \geq 2$$ and $$x_1x_2 \geq 1$$.

We get $$-\frac{k}{2} \geq 2$$ or equivalently $$k \leq -4$$ and $$-\frac{k+1}{2}\geq 1$$ or equivalently $$k \leq -3$$. Together they give you $$k \leq -4$$.

You have to check one last thing: you equation must have two real roots, so $$\Delta \geq 0$$. Here $$\Delta=k^2+4\cdot 2 (k+1)=k^2+8k+8$$. Now $$k^2+8k+8 \geq 0$$ and we have that $$k^2+8k+8=(k-2\sqrt{2}+4)(k+2\sqrt{2}+4)$$ so $$k \leq -2\sqrt{2}-4$$ or $$k \geq 2\sqrt{2}-4$$.

Now we can combine our conditions and the final answer is $$k \in (-\infty, -2\sqrt{2}-4]$$

• Yes but 3 doesn't satisfy it. Neither does the other element of the 2nd set. Neither does -7 , which is supposed to satisfy it. I did all you did. But I think I'm doing a mistake because the answer isn't satisfying. – NightKruger Jun 15 at 18:46
• @NightcoRohak I made a stupid mistake at the beginning: $-\frac{k}{2} \geq 2$ implies $k \leq -4$ and $-\frac{k+1}{2} \geq 1$ implies $k \leq -3$ – user289143 Jun 15 at 18:55
• Oh sorry , 7 does satisfy. – NightKruger Jun 15 at 19:37

HINT: Roots of the given quadratic equation are: $$x_{1,2} = \frac{-k\pm \sqrt{k^2+4\cdot2(k+1)}}{2\cdot2} = \frac{-k\pm \sqrt{k^2+8k+8}}{4}$$ and we are asked to find interval for $$k$$ when both of these roots are greater than $$1$$. After Dr. Sonnhard Graubner's comment, I realized that you should also check for $$k^2+8k+8 \ge 0$$. This will give you one more interval for $$k$$. Can you take it from here?

• Do you mean real roots?Since $$k^2+8k+16+8-16=(k+4)^2-8$$ – Dr. Sonnhard Graubner Jun 15 at 18:28
• Quadratic formula can find the complex roots as well. But in this case, it would be weird for a complex number with non-zero imaginary part to be greater than $1$. Therefore we might have to find an interval for roots to be real first. – ArsenBerk Jun 15 at 18:33

Let $$f(x)=2x^2+kx-(k+1)$$. The parabola $$y=f(x)$$ is concave up (because the coefficient of $$x^2$$ is positive). For the roots to be greater than $$1$$, we need three things:

real roots, $$f(1)>0$$ and the $$x-$$coordinate of the vertex of the parabola to be greater than $$1$$.

1). $$k^2+8(k+1) \geq 0 \implies \color{blue}{k \geq -4+2\sqrt{2} \text{ or } k \leq -4-2\sqrt{2}}$$.

2). $$f(1)=1>0$$. So nothing to be done here.

3). The vertex occurs at $$x=-\frac{k}{4}$$. So we want $$\frac{-k}{4}>1 \implies \color{blue}{k <-4}$$.

The common values of $$k$$ are $$\color{red}{k \leq -4-2\sqrt{2}}$$

• You put the wrong $c$ while computing the discriminant – user289143 Jun 15 at 18:59
• @user289143 Thanks for pointing it out. Fixed the typo. – Anurag A Jun 15 at 19:01

Denote $$p(x)=2x^2+kx-(k+1)$$.

• First $$p(x)$$ must have real roots, i.e. its discriminant $$\;\Delta(k)=k^2+8(k+1)\ge 0$$. Its roots are $$\kappa_1=-4-2\sqrt 2,\quad \kappa_2=-4+2\sqrt. 2$$ So the first condition is $$\;k\in(-\infty, \kappa_1]\cup[\kappa_2,+\infty)$$.
• Observe that $$\;\Delta(1)>0$$, so $$1$$ is outside the the interval of the roots ($$\xi_1,\xi_2$$), i.e. we have either $$1<\xi_1\le\xi_2$$ or $$\xi_1\le \xi_2<1$$.
• We want to have the first case. This is equivalent to $$1<\frac{\xi_1+\xi_2}2=-\frac k4\iff k<-4.$$ Grouping these conditions, we obtain $$k\le -4-2\sqrt 2.$$

If $$k> 0$$ then for $$x> 1$$ $$\therefore\,2\,x^{\,2}+ kx- k- 1> 2+ k- k- 1> 0$$ If $$-\,4\leqq k\leqq 0$$ then for $$x> 1$$ $$\therefore\,2\,x^{\,2}+ kx- k- 1\geqq 2\,x^{\,2}- 4(\,x- 1\,)- 1> 0$$ But $$2\,x^{\,2}+ kx- k- 1= 0\,\therefore\,k< -\,4$$.

Using $$\lceil$$ https://en.wikipedia.org/wiki/Quadratic_function $$\rfloor$$ we receive $$2\,x^{\,2}+ kx- k- 1= 0$$ has two roots then $$\therefore\,{\rm discriminant}= k^{\,2}+ 8\,k+ 8> 0$$ $$\therefore\,(\,k+ 4\,)^{\,2}> 8\,\therefore\,k< -\,4- \sqrt{8}$$

$$(x_1-1)(x_2-1)>0 \implies x_1x_2-(x_1+x_2)+1>0$$

so by Vieta formulas we have $$-{k+1\over 2}+{k\over 2} +1>0\implies k\in \mathbb{R}$$

So you have to check only when the disciriminat is positive and you are done.