Quadratic inequality with Modulus of a Linear Equation

Adolfo Hebert

Adolfo Hebert

Answered question

2022-03-31

Quadratic inequality with Modulus of a Linear Equation : Solution Verification & Clarification Needed
x2|5x3|x<2
For this problem I took two cases where in one case 5x30 and in second case 5x3<0
1) 5x30:
2) 5x3<0

Answer & Explanation

Abdullah Avery

Abdullah Avery

Beginner2022-04-01Added 19 answers

Let me forcus on when 5x3<0, we have x<35 and x2|5x3|x<2
Becomes x2(35x)x<2
x2+4x5<0
(x+5)(x1)<0
Hence we have 5<x<1 and x<35, hence taking intersection, we have 5<x<35.
To answer the question, why did we exclude x=5, consider (x+5)(x1)<0, if we substute x=5 inside, we get 0<0 which is not true.

Esteban Sloan

Esteban Sloan

Beginner2022-04-02Added 21 answers

If 5x30, we have x35 and
x2|5x3|x<2
becomes x25x+3x<2
x26x+1<0
63642<x<6+3642
322<x<3+22
Taking intersection with x35, we have
35x<3+22

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