For which $x \in \mathbb{R}$ is the inequality $x^2-5x - 6\ge 0$ true?

Question

The inequality is: $$x^{2}-5x - 6 \geq 0. $$

I know this sounds like a stupid question, but via just guessing and checking I got, that $$3 \leq x \leq 0$$

but I am unsure how to formulate a proper proof

Answer 1

Factorize the quadratic: we have $x^2-5x-6=(x-6)(x+1)$. If $a$ and $b$ are real numbers, then $ab\ge0$ if and only if either both $a,b\ge0$ or $a,b\le0$. So $x^2-5x-6\ge0$ if $x-6,x+1\ge0$ or $x-6,x+1\le0$, or in other words, if $x\ge6$ or if $x\le-1$.

Answer 2

$x^2-5x-6=(x-6)(x+1)\geq 0$

Then both $(x-6),(x+1)\geq 0$ or $(x-6),(x+1)\leq 0$ i.e $x\in [6,\infty)$ or $x\in (-\infty ,-1]$

Answer 3

Use the quadratic equation formulas and factor the terms as:

$$(x-\rho_1)(x-\rho_2) \ge 0$$

(assume $\rho_1 \le \rho_2$)

Then for this product to be greater or equal to zero we have:

1. $x=\rho_1$ or $x=\rho_2$ (equal to zero)
2. or each term must be positive, i.e $x > \rho_2$
3. or each must be negative. i.e $x < \rho_1$

so all in all

$$ x \le \rho_1$$

or

$$x \ge \rho_2$$

solve and substitue $\rho_1, \rho_2$ with the actual roots and there it is!

Answer 4

First factorize the quadratic equation, or solve it. By factorizing you will get (x-6)(x+1). So at x=6,-1 the quadratic will give an value zero.

Shortcut: Now just see whether the sign of the second degree term is positive or negative. If its positive then the parabola of the quadratic opens upwards and hence the quadratic will have <0 values between -1 and 6, (if 2nd degree term is negative which is not in this case, the parabola will open downwards and the inequality would be true only between -1 and 6.)

Now you want to know where the this equation is greater than or equal to zero. (We know where the equation is equal to zero at 6,-1)

To know where the equation is greater than zero, just take out the range of values for which the equation is less than zero, and remove them from the domain. Your domain where the inequality is true is ready.

Now to know where the equation goes below zero, its very easy just differentiate and you will know where the inequality decreases and increases.(You already know where the equation is =0. So now just draw a graph which will be a parabola opening upwards intersecting the x axis at -1 and 6. From the parabola it would be very clear were the equation is less than zero.

(-1,6)_domain where the inequality will not be valid since the quadratic will be less than zero. So for R-(-1,6) this inequality is valid.