Find-the-solution-set-of-inequality-3x-4-x-2-

Question Number 153070 by bobhans last updated on 04/Sep/21

F i n d t h e s o l u t i o n s e t o f i n e q u a l i t y

\sqrt{3 x + 4} ⩾ x - 2

Answered by Rasheed.Sindhi last updated on 04/Sep/21

\sqrt{3 x + 4} ⩾ x - 2

{(\sqrt{3 x + 4} ⩾ x - 2)}^{2}

3 x + 4 ⩾ x^{2} - 4 x + 4

x^{2} - 7 x ⩽ 0

x (x - 7) ⩽ 0

(x ⩽ 0 \land x - 7 ⩾ 0) \lor (x ⩾ 0 \land x - 7 ⩽ 0)

(x ⩽ 0 \land x ⩾ 7) \lor (x ⩾ 0 \land x ⩽ 7)

0 ⩽ x ⩽ 7

Commented by MJS_new last updated on 04/Sep/21

\sqrt{3 x + 4} defined for x ⩾ - \frac{4}{3}

\sqrt{3 x + 4} = x - 2 \Rightarrow only one solution (squaring

leads to false solutions) : x = 7

\Rightarrow

the inequality holds for - \frac{4}{3} ⩽ x ⩽ 7

Commented by Rasheed.Sindhi last updated on 04/Sep/21

THANKS Sir! I think Ineed more

study regarding the topic .

Answered by john_santu last updated on 04/Sep/21

{\begin{cases} x ⩾ - \frac{4}{3}; w h e n x - 2 < 0 w e g e t s o l u t i o n - \frac{4}{3} ⩽ x < 2 \\ w h e n x - 2 ⩾ 0; x ⩾ 2 \Rightarrow s q u a r i n g b o t h s i d e s \end{cases}

\Rightarrow 3 x + 4 ⩾ x^{2} - 4 x + 4

\Rightarrow x^{2} - 7 x ⩽ 0; 0 ⩽ x ⩽ 7 w e g e t s o l u t i o n 2 ⩽ x ⩽ 7

T h e r e f o r e t h e s o l u t i o n s e t i s

\Rightarrow x \in [- \frac{4}{3}, 7]

Answered by talminator2856791 last updated on 04/Sep/21

0 ⩾ x - 2 - \sqrt{3 x + 4}

0 ⩾ ((x - 2) - \sqrt{3 x + 4}) ((x - 2) + \sqrt{3 x + 4})

0 ⩾ {(x - 2)}^{2} - (3 x + 4)

0 ⩾ x^{2} - 4 x + 4 - 3 x - 4

0 ⩾ x^{2} - 7 x

0 ⩾ x (x - 7)

0 ⩽ x ⩽ 7

Commented by MJS_new last updated on 04/Sep/21

0 ⩾ ((x - 2) - \sqrt{3 x + 4}) ((x - 2) + \sqrt{3 x + 4})

if (x - 2) + \sqrt{3 x + 4} ⩾ 0 \Leftrightarrow x ⩾ 0

0 ⩾ ((x - 2) - \sqrt{3 x + 4}) ((x - 2) + \sqrt{3 x + 4})

if (x - 2) + \sqrt{3 x + 4} < 0 \Leftrightarrow - \frac{4}{3} ⩽ x < 0

anyway you cannot solve it this way

y_{1} = x - 2 is a straight line

y_{2} = \sqrt{3 x + 4} is a half parabola

how ever you reach 3 x + 4 you get false solutions

\sqrt{3 x + 4} is defined for x ⩾ - \frac{4}{3}

x - 2 = \sqrt{3 x + 4} has only 1 solution : x = 7

{it}^{'} s easy to show that

x - 2 < \sqrt{3 x + 4} for - \frac{4}{3} ⩽ x < 0

and from both x - 2 and \sqrt{3 x + 4} strictly

increasing it follows that x - 2 > \sqrt{3 x + 4} for

x > 7

\Rightarrow solution is - \frac{4}{3} ⩽ x ⩽ 7

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