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

How do you solve the inequality ${x}^{2}-6x-7<0$
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Elaine Verrett
Explanation:
First, factorise the x terms.
${x}^{2}-6x-7<0$
$\left(x-7\right)\left(x+1\right)<0$
To find the roots:
$x-7=0$ or $x+1=0$
$x=7$ or $x=-1$
The question asks us to find the portion of the curve which is less than zero $\left(<0\right)$. 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.
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 $-1
(If the question states ${x}^{2}-6x-7>0$, the portions of the curve that are positive is when $x<-1$ and $x>7$)
###### Not exactly what you’re looking for?
Jenny Bolton
Explanation the quadratic on the left side
$⇒\left(x-7\right)\left(x+1\right)<0$
find the zeros
$x=-1$ and $x=7$
These indicate where the function changes sign
the zeros split the x-axis into 3 intervals

Consider a test point in each interval
we want to find where the function is negative, <0
substitute each tst point into the function and consider its