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# How to solve the following quadratic inequalities and write them in the interval notation?

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How to solve the following quadratic inequalities and write them in the interval notation?

x2 < 25

(2x-3)(2x+5) > -7

Jun 8, 2019

#1
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x^2 < 25

x^2 - 25 < 0

Factor

(x + 5) ( x - 5) < 0

Set each factor to 0   and solve

x + 5  =  0            x  -  5  = 0

x = - 5                  x   = 5

We have three possible intervals that solve this...... (-inf, -5)  or (-5, 5)  or (5, 0)

What happens is that either the two outside intervals solve the original equation or the middle interval does

Test x =  0,  [which is a point in the middle interval ] in the original inequality

0^2 < 25    which is true

So....the interval  that solves the inequality  is  (-5, 5).....we don't include the endpoints because of the "<" sign

Here's a graph : https://www.desmos.com/calculator/fr1w7y9y6f   Jun 8, 2019
#2
+2

(2x-3)(2x+5) > -7   simplify

4x^2 - 6x + 10x - 15 > -7

4x^2 + 4x - 15 > -7        add 7 to both sides

4x^2 + 4x - 8 > 0         divide through by 4

x^2 + x - 2 >  0           factor

(x + 2) ( x - 1) > 0

As before.....set each factor to 0 and solve

x + 2  = 0             x - 1  = 0

x = -2                      x  = 1

We have three intervals ..... ( -inf, -2) or (-2, 1)  or  (1, inf)

Test x = 0  in the original inequality  and we get that

(2(0)-3)(2(0)+5) > -7

(-3) (5) > -7

-15 > -7

This is not true

So....the intervals that solve  this   are   (-inf, -2)  and ( 1, inf)

Here's the graph : https://www.desmos.com/calculator/7dgbgkrmaw   Jun 8, 2019