# How do you solve the inequality x^2-6x-7<0?

Jun 10, 2017

$- 1 < x < 7$

#### Explanation:

First, factorise the x terms.

${x}^{2} - 6 x - 7 < 0$

$\left(x - 7\right) \left(x + 1\right) < 0$

To find the roots:

$x - 7 = 0 \mathmr{and} x + 1 = 0$

$x = 7 \mathmr{and} x = - 1$

The question asks us to find the portion of the curve which is less than zero (<0).
Thus, we need to show by giving a range of values of $x$.

Do a rough sketch of a graph showing
- ${x}^{2}$ curve on x and y axis
- curve must be positive, meaning it has a minimum point. (like a smile)
- label the x-intercepts with the values found above.

This is a zoomed in graph
graph{x^2-6x-7 [-6.875, 13.125, -4.96, 5.04]}

This next step is a marking point.
Recall when drawing a graph, you have to mark an "x" or "+" at each coordinates. This is different when involved with inequalities.

• When the equality is: $< , >$, you have to draw a circle around the x-intercept, instead of marking an "x".

• When the equality is: $\le , \ge$, you have to shade a dot at the x-intercept, instead of marking an "x".

Since the equality given is "<", draw a small circle at the intercepts.

The portion of the curve that is negative falls below $y = 0$

This means the curve is negative between $x = - 1$ and $x = 7$

Which is expressed as $\textcolor{red}{- 1 < x < 7}$

(If the question states ${x}^{2} - 6 x - 7 > 0$, the portions of the curve that are positive is when $x < - 1 \mathmr{and} x > 7$)

Jun 10, 2017

$- 1 < x < 7$

#### Explanation:

$\text{factorise the quadratic on the left side}$

$\Rightarrow \left(x - 7\right) \left(x + 1\right) < 0$

$\text{find the zeros}$

$x = - 1 \text{ and } x = 7$

$\text{these indicate where the function changes sign}$

$\text{the zeros 'split' the x-axis into 3 intervals}$

$x < - 1 , \textcolor{w h i t e}{x} - 1 < x < 7 , \textcolor{w h i t e}{x} x > 7$

$\text{consider a "color(blue)"test point "" in each interval}$

$\text{we want to find where the function is negative } , < 0$

$\text{substitute each test point into the function and }$
$\text{consider it's sign}$

$\textcolor{red}{x = - 2} \to \left(-\right) \left(-\right) \to \textcolor{red}{\text{ positive}}$

$\textcolor{red}{x = 2} \to \left(-\right) \left(+\right) \to \textcolor{b l u e}{\text{ negative}}$

$\textcolor{red}{x = 10} \to \left(+\right) \left(+\right) \to \textcolor{red}{\text{ positive}}$

$\Rightarrow - 1 < x < 7 \text{ is the solution}$