x^2-4x-45>0 solve for x values

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on


First we will factorize:


That means:

x-9>0     and    x+5>0

==> x>9  and x>-5

Then x belongs to the interval (9,infinity)....(1)


x-9<0    and x+5<0

x<9    and    x< -5

Then x belongs to the interval (-infinity, -5)....(2)

From (1) and (2):

x belongs to the interval (-inf, -5)U(9,inf)

OR x = R-[-5, 9]


neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To solve x^2-4x-45 > 0.


The principle  of solving the quadratic inequality like this( expressed as greater than zero) is  finding the roots of the quadratic equation and see that the quadratic expression  is positive only if the x values are outside the  interval representented by the roots. This involves in factorising the quadratic expression and see that values of the variable x makes (i) both factors are positive or greater than zero, or (ii) both factors less than zero, so that their product is positive.

So x^2-4x-45 > 0. We factorise the left.

x^2-9x+5x-45 > 0

x(x-9) +5(x-9) > 0. Pull out (x-9).

(x-9)(x+5) > 0.

So for values of x < -5 makes both factors negative .So x< -5 makes the product positive.

Similarly for values of x > 9 both factors are positive. And so, the product of the facrors is positive.

Therefore, in the interval notation,  when x belongs to (-infinity , -5) U (9 , infinity )  the given quadratic inequality holds good.


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