solve and check (x+2)(x+5)>0

Question

Quadratic inequalities.

Answer 1

Given quadratic inequality (x + 2)(x + 5) > 0

x^2+ 5x + 2x +10 > 0

x^2 + 7x + 10 > 0

Firstly let us find where it is equal to zero.

x^2 + 7x + 10  = 0

Compare it to quadratic equation ax^2 + bx + c  = 0.

a = 1, b = 7, c = 10.

x  = [-b±√(b^2-4ac)]/2a

x  = [-7±√(49-40)]/2

x  = [-7±√9]/2

x  = [-7±3]/2

x  = [-7+3]/2 and x = [-7-3]/2

x  = -4/2 and x = -10/2

x  = -2 and x = -5

Solution x  <  -2 or x  < -5.

Check :

If x  = -6

x ^2 + 7x + 10  >  0

36 - 42 +10 > 0

4 > 0

Above stament is true.

Answer 2

Given quadratic inequality (x + 2)(x + 5) > 0

x^2+ 5x + 2x +10 > 0

x^2 + 7x + 10 > 0

Firstly let us find where it is equal to zero.

x^2 + 7x + 10  = 0

Compare it to quadratic equation ax^2 + bx + c  = 0.

a = 1, b = 7, c = 10.

x  = [-b ± √(b^2 - 4ac)]/2a

x  = [-7 ± √(49 - 40)]/2

x  = [-7 ± √9]/2

x  = [-7 ± 3]/2

x  = [-7 + 3]/2 and x = [-7 - 3]/2

x  = - 4/2 and x = - 10/2

x  = -2 and x = -5

So between -2 and -5 the function will be either

always greater than zero or less than zero.

Let's pick a value in between and test it.

At x = -3

(x + 2)(x + 5) = (- 3 + 2)(- 3 + 5) = - 1(2) = -2

So between -2 and -5 the function less than zero.

(x + 2)(x + 5) > 0 on the interval (-∞,-5)  and (-2,∞)

Solution x > -2 and x < -5

Check

For x  = 2

(x + 2)(x + 5) > 0

4(7) > 0

28 > 0

Above statement is true.

For x  = -4

(x + 2)(x + 5) > 0

-2(1) > 0

-2 > 0

Above statement is false.