How to solve for (x + 5)(x - 5) > 0?

Question

Solve for x (real number).

Answer 1

If a<b and (x-a)(x-b) >0 the x<a or b<x.

So here x<-5 or 5<x. Simple

You may draw a graph of this and check.

Answer 2

what you want to do is FOIL (first, outside, inside, last) the left side first. that way, you'll get x^2+5x-5x-25>0.

that simplifies to x^2-25>0.

add 25 to both sides, to cancel it out on the left. x^2-25+25>0+25

you now have x^2>25

find the square root of both sides. x>+5 or x<-5.

you'd write that as -5<x<5.

hope i helped!

Tunz

Answer 3

x² - 25 > 0

x² > 25

x > 5 or x < -5

Answer 4

If two numbers multiply to come out greater than zero (i.e. positive) then either both are positive or both are negative.  That happens here if x > 5 (both will be positive) or x < -5 (both will be negative).

So x>5 or x<-5

Answer 5

sketch the parabola and find where it is greater than 0

Its a positive parabola with x intercepts at 5 and -5

so (x + 5)(x - 5) > 0 when x > 5 and x < -5

In interval notation that's (-∞, -5) U (5, ∞)

Answer 6

this happens if both the factor are pos or both neg

i.e

1) x+5>0 & x-5>0

both true for x>5

2) x+5<0 & x-5<0

both true for x<-5

hence ans (solution set) : (-inf,-5) U (union) (5,inf)

Answer 7

(negative infinity, -5) U (+5, positive infinity)

Answer 8

x>5 and x<-5