Question

(x-6)(x+7) > 0

Answer 1

use a coordinate lane

Answer 2

do you understand the sings I put on the lane

Answer 3

or you could just solve x>6 x>-7

Answer 4

you need values of x, where there stands + on line. so it will be x<-7 and x>6

Answer 5

isnt it the other way x>-7

Answer 6

your subtracting, not dividing or multiplying so the signs dont switch

Answer 7

you must have mistaken, for example x=0 and the product won't be bigger than 0

Answer 8

you need values of x where factors are both - or both +

Answer 9

x-6=0
x=-6
x+7=0
x=-7
\[\left( x-a \right)\left( x-b \right)>0\]
means x does not lie between a and b.
\[\left( x-a \right)\left( x-b \right)<0\]
means x lies between a and b

Answer 10

but when solving for an equation, you get each parenthesis and = it to 0. so isn't it x-6>0 and x+7>0

Answer 11

the @surjithayer is correct. I didn't think of that method.

Answer 12

step by step solution attached below

Answer 13

one more method
\[\left( x-6 \right)\left( x+7 \right)>0\]
\[x^2+x-42>0\]
adding 42 to both sides
\[x^2+x>42\]
\[adding~both~sides~\left( \frac{ co-efficient~of~x }{ 2 } \right)^2~i.e.,\left( \frac{ 1 }{ 2 } \right)^2~or~\frac{ 1 }{ 4 }\]
\[x^2+x+\frac{ 1 }{ 4 }>42+\frac{ 1 }{ 4 }\]
\[\left( x+\frac{ 1 }{ 2 } \right)^2>\frac{ 168+1 }{ 4 }\]
\[\left( x+\frac{ 1 }{ 2 } \right)^2>\left( \frac{ 13 }{ 2 } \right)^2\]
\[\left| x+\frac{ 1 }{ 2 } \right|>\frac{ 13 }{ 2 }\]
\[\left| x+a \right|>b~\rightarrow~x+a <-b~and~x+a >b\]
complete it.