Solve-a-x-3-2-gt-5-b-3x-2-gt-12-

Question Number 80405 by TawaTawa last updated on 02/Feb/20

Solve :

(a) {(x - 3)}^{2} > - 5

(b) {3 x}^{2} > - 12

Commented by MJS last updated on 02/Feb/20

a {(x - 3)}^{2} ⩾ 0 > - 5 \Rightarrow true for all x \in R

b 3 x^{2} ⩾ 0 > - 12 \Rightarrow true for all x \in R

strange examples \dots

Commented by TawaTawa last updated on 02/Feb/20

Sir, no workings ? ?

Commented by MJS last updated on 02/Feb/20

no workings .

\forall r \in R : r^{2} ⩾ 0

nothing else is necessary

Commented by TawaTawa last updated on 02/Feb/20

God bless you sir .

Commented by sandy_delta last updated on 03/Feb/20

h o w a b o u t c o m p l e x n u m b e r ?

Commented by jagoll last updated on 03/Feb/20

(a) {(x - 3)}^{2} + 5 > 0

(x - 3 - i \sqrt{5}) (x - 3 + i \sqrt{5}) > 0

x < 3 - i \sqrt{5} \lor x > 3 + i \sqrt{5}

Commented by jagoll last updated on 03/Feb/20

(b) 3 x^{2} + 12 > 0

3 (x^{2} + 4) > 0

3 (x + 2 i) (x - 2 i) > 0

x < - 2 i \lor x > 2 i

Commented by MJS last updated on 03/Feb/20

complex numbers cannot be sorted using

< or > so this makes no sense at all

Commented by jagoll last updated on 03/Feb/20

- 2 i < 2 i s i r ?

Commented by MJS last updated on 03/Feb/20

- 2 i < 2 i ∣ - 2 i

- 4 i < 0 ∣ \div (- 4)

i > 0

we are allowed to multiply both sides with

a number > 0 without changing the <> signs

i > 0 ∣ \times i

i^{2} > 0

- 1 > 0

wrong

i < 0 ? ? ?

we must change the <> signs when multiplying

with a number < 0

i < 0 ∣ \times i

i^{2} > 0

- 1 > 0

wrong

\Rightarrow i i s n e i t h e r > 0 n o r < 0

Commented by jagoll last updated on 03/Feb/20

t h a n k y o u m i s t e r

Answered by sandy_delta last updated on 03/Feb/20

(b) 3 x^{2} > - 12

x^{2} > - 4

x^{2} - (- 4) > 0

(x + \sqrt{- 4}) (x - \sqrt{- 4}) > 0

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

x < - 2 i \lor x > 2 i

Commented by MJS last updated on 03/Feb/20

see my above comment and / or please tell

me if

- 3 i + 5 is smaller or greater than - 2 i

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