# x^2 -6x < -5

hala718 | Certified Educator

x^2 -6x < -5

x^2 -6x + 5 < 0

Now factorize:

(x-5)(x-1) <0

In order for the inequality to hold, we have two options:

option 1:

x-5 <0    and  x-1 >0

x<5    and x >1

==> 1<x<5

==> x belongs to the interval (1,5)

options 2:

x-5>0     and    x-1 <0

x> 5     and    x < 1

The answer here is a null set because  the intersection is an empty set.

x belongs to (1,5)

giorgiana1976 | Student

First of all, let's solve the expression like an equation. For this reason, we'll put:

x^2 -6x + 5 = 0

x1 = [6 + sqrt(36-20)]/2

x1 = (6+4)/2

x1 = 5

x2  =(6-4)/2

x2 = 1

Now, we'll follow the rule: between the solution of the equation, the expression will have the opposite sign of the coefficient of x^2, that means that x^2 -6x + 5 < 0, outside the solutions, the expression will be positive.

So, the conclusion is that the expression is negative for values of x which are in the interval (1,5).

neela | Student

To solve x^2-6x < -5

We rewrite the inequation as:

x^2-6x +5 < 0

x^2-6x+5 < 0

x^2-5x-x+5 < 0

x(x-5)-1(x-5) < 0

(x-5)(x-1) < 0

The expression (x-5)(x-1) is negative or both factors with opposite signs when x belongs (1,5)

neela | Student

x^2-6x <-5

To solve this quadratic inequality we write it as

x^2-6x+5  < 0.

We know that if x1 and x2 are the real roots then x^2-6x+5 < o for x belonging to the interval  (x1 , x2)

But  x^2-6x+5 <0

x^2-5x-x+5 < 0

x(x-5)-1(x-5) < 0

(x-5)(x-1) < 0

x1 = 1  and x2 = 5. are the roots of x^2-6x+5 = 0.

So x belongs to (x1 , x2). Or

x belongs to (1,5)