Question

solve inequality in quadratic equations.
| x-3 |/|x^2-4| < = 1
find possible real value of x.

Answer 1

Heya !!

ι ∂σ тнє αиѕωєя ιи ∂єтαιℓ. ѕιмρℓу ωαу.

gινєи |x-3|/|x²-4| ≤ 1

¢ℓєαяℓу x ≠ 2 or -2.

C͟͟a͟͟s͟͟e͟͟❶ : |x| < 2, ie, -2 < x < 2

So numerator is negative
Denominator is negative.
= (3 - x)/(4 - x²) ≤ 1

= 4 - x²+x - 3 ≥ 0

= x² - x -1 ≤ 0

=> Roots = (1 ± √5)/2 valid as it is in
the range |x| < 2
=> so (1 - √5)/2 ≤ x ≤ (1 + √5)/2

======

C͟͟a͟͟s͟͟e͟͟❷: 2 < x < 3

(3 -x)/(x²- 4) ≤ 1
x² - 4 + x -3 ≥ 0
x² + x -7 ≥ 0
roots : = [-1 ± √29] 12 ,(-1 - √29)/
2 is out of range.
=> 2 < x < [ √29 - 1]/2

C͟͟a͟͟s͟͟e͟͟❸: x > 3

x - 3 ≤ x² -4
x² - x - 1 ≥ 0
roots : (1 ± √5)/2
so valid range : x ≥ (√5 + 1)/2

C͟͟a͟͟s͟͟e͟͟❹ : ∞ < x < -2

(3 - x) ≤ (x² - 4)
x² - 7 + x ≥ 0
roots : (-1 ± √29)/2
so x<(-1 -√29)/2 or
x > (√29 - 1)/2

checking with the range of case4, we
get -8,< x< (-1 -√29)/2

ᏩᏞᎪᎠ ᎻᎬᏞᏢ YᎾᏌ.
ᏆᎢ ᎻᎬᏞᏢᏚ YᎾᏌ
ᎢᎻᎪNK YᎾᏌ☻
@vaibhav246

Answer 2

Step-by-step explanation:

thanks for explanation