What's the solution to2|2x−a|&lt;|x+3a|(found critical values)

Step 1An inequality|P(x)|&lt;|Q(x)|is the same as(P(x))2&lt;(Q(x))2which becomes(Q(x)−P(x))(Q(x)+P(x))&gt;0In your case you get(x+3a−4x+2a)(x+3a+4x−2a)&gt;0and, in shorter form,(3x−5a)(5x+a)&lt;0Now it’s easier: for a=0 you have no solution. For a&gt;0, the roots are −a5&lt;5a3, so−15a&lt;x&lt;53aYou can do the a&lt;0 case.

Step 1For a=0 our inequality has no solutions.But for a≠0 it's equivalent to4(2x−a)2&lt;(x+3a)2or15x2−22ax−5a2&lt;0or11a−14|a|15&lt;x&lt;11a+14|a|15.

Step 1An inequality|P(x)|&lt;|Q(x)|is the same as(P(x))2&lt;(Q(x))2which becomes(Q(x)−P(x))(Q(x)+P(x))&gt;0In your case you get(x+3a−4x+2a)(x+3a+4x−2a)&gt;0and, in shorter form,(3x−5a)(5x+a)&lt;0Now it’s easier: for a=0 you have no solution. For a&gt;0, the roots are −a5&lt;5a3, so−15a&lt;x&lt;53aYou can do the a&lt;0 case.

Step 1For a=0 our inequality has no solutions.But for a≠0 it's equivalent to4(2x−a)2&lt;(x+3a)2or15x2−22ax−5a2&lt;0or11a−14|a|15&lt;x&lt;11a+14|a|15.

Check the solution to "What's the solution to2|2x−a|<|x+3a|(found critical values)"

High School
Algebra
Algebra I
Quadratic function and equation

hinnikendvjg

Answered question

2022-03-14

What's the solution to

2 | 2 x - a | < | x + 3 a |

(found critical values)

Answer & Explanation

InnovaRat0r5

Beginner2022-03-15Added 2 answers

Step 1
An inequality

| P (x) | < | Q (x) |

is the same as

{(P (x))}^{2} < {(Q (x))}^{2}

which becomes

(Q (x) - P (x)) (Q (x) + P (x)) > 0

In your case you get

(x + 3 a - 4 x + 2 a) (x + 3 a + 4 x - 2 a) > 0

and, in shorter form,

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

Now it’s easier: for

a = 0

you have no solution. For

a > 0

, the roots are

- \frac{a}{5} < \frac{5 a}{3},

- \frac{1}{5} a < x < \frac{5}{3} a

You can do the

a < 0

case.

Pamela Browning

Beginner2022-03-16Added 7 answers

Step 1
For

a = 0

our inequality has no solutions.
But for

a \neq 0

it's equivalent to

4 {(2 x - a)}^{2} < {(x + 3 a)}^{2}

15 x^{2} - 22 a x - 5 a^{2} < 0

\frac{11 a - 14 | a |}{15} < x < \frac{11 a + 14 | a |}{15} .

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