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Question Number 178612 by Acem last updated on 19/Oct/22

Be calm then solve ∣((x^2 +7x−8)/(x+3))∣≥ 2

$${Be}\:{calm}\:{then}\:{solve}\:\mid\frac{{x}^{\mathrm{2}} +\mathrm{7}{x}−\mathrm{8}}{{x}+\mathrm{3}}\mid\geqslant\:\mathrm{2} \\ $$

Answered by cortano1 last updated on 19/Oct/22

 x≠−3   ∣x^2 +7x−8∣≥∣2x+6∣   (x^2 +7x−8+2x+6)(x^2 +7x−8−2x−6)≥0   (x^2 +9x−2)(x^2 +5x−14)≥0    [(x+(9/2))^2 −((89)/4)−](x+7)(x−2)≥0   (x+((9+(√(89)))/2))(x+(((9−(√(89)))/2)))(x+7)(x−2)≥0

$$\:\mathrm{x}\neq−\mathrm{3} \\ $$$$\:\mid\mathrm{x}^{\mathrm{2}} +\mathrm{7x}−\mathrm{8}\mid\geqslant\mid\mathrm{2x}+\mathrm{6}\mid \\ $$$$\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{7x}−\mathrm{8}+\mathrm{2x}+\mathrm{6}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{7x}−\mathrm{8}−\mathrm{2x}−\mathrm{6}\right)\geqslant\mathrm{0} \\ $$$$\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{9x}−\mathrm{2}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{5x}−\mathrm{14}\right)\geqslant\mathrm{0} \\ $$$$\:\:\left[\left(\mathrm{x}+\frac{\mathrm{9}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{89}}{\mathrm{4}}−\right]\left(\mathrm{x}+\mathrm{7}\right)\left(\mathrm{x}−\mathrm{2}\right)\geqslant\mathrm{0} \\ $$$$\:\left(\mathrm{x}+\frac{\mathrm{9}+\sqrt{\mathrm{89}}}{\mathrm{2}}\right)\left(\mathrm{x}+\left(\frac{\mathrm{9}−\sqrt{\mathrm{89}}}{\mathrm{2}}\right)\right)\left(\mathrm{x}+\mathrm{7}\right)\left(\mathrm{x}−\mathrm{2}\right)\geqslant\mathrm{0} \\ $$

Commented by Acem last updated on 19/Oct/22

 Good sir! try to find the required x values    Tip: ∣P ∣≥ d ; d≥ 0 ⇒     Either P ≥ d     Or     P ≤ −d

$$\:{Good}\:{sir}!\:{try}\:{to}\:{find}\:{the}\:{required}\:{x}\:{values} \\ $$$$ \\ $$$$\boldsymbol{{Tip}}:\:\mid{P}\:\mid\geqslant\:{d}\:;\:{d}\geqslant\:\mathrm{0}\:\Rightarrow \\ $$$$ \\ $$$$\:\boldsymbol{{Either}}\:\boldsymbol{{P}}\:\geqslant\:\boldsymbol{{d}}\:\:\:\:\:\boldsymbol{{Or}}\:\:\:\:\:\boldsymbol{{P}}\:\leqslant\:−\boldsymbol{{d}} \\ $$

Commented by cortano1 last updated on 19/Oct/22

i don′t try it

$$\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{try}\:\mathrm{it} \\ $$

Commented by Acem last updated on 19/Oct/22

Ok friend, you can check it below

$${Ok}\:{friend},\:{you}\:{can}\:{check}\:{it}\:{below} \\ $$

Answered by Acem last updated on 19/Oct/22

  Either ((x^2 +7x−8)/(x+3))≥ 2 ∣_A    Or    ((x^2 +7x−8)/(x+3))≤ −2 ∣_B     A: (((x−2)(x+1))/(x+3))≥ 0        x       −∞           −7          −3            2           +∞  (x−2)(x+1)                +     0      −            −     0     +      x+3                 −             −     0     +            +     Frac.                −    0      +      ∥     −    0     +     Either x∈ [−7, −3[ ∪ [2, +∞[    B:  (((x+(9/2))^2 −((89)/4))/(x+3))≤ 0    x= ((−9− (√(89)))/2) ≅ −9.21  or x= ((−9+ (√(89)))/2) ≅ 0.21          x       −∞             x_1           −3            x_2            +∞  (x−2)(x+1)                +     0      −            −     0     +      x+3                 −             −     0     +            +     Frac.                −    0      +      ∥     −    0     +     Or x∈ ]−∞, ((−9− (√(89)))/2) ] ∪ ]−3, ((−9+ (√(89)))/2) ]

$$ \\ $$$${Either}\:\frac{{x}^{\mathrm{2}} +\mathrm{7}{x}−\mathrm{8}}{{x}+\mathrm{3}}\geqslant\:\mathrm{2}\:\mid_{\boldsymbol{{A}}} \:\:\:{Or}\:\:\:\:\frac{{x}^{\mathrm{2}} +\mathrm{7}{x}−\mathrm{8}}{{x}+\mathrm{3}}\leqslant\:−\mathrm{2}\:\mid_{\boldsymbol{{B}}} \\ $$$$ \\ $$$$\boldsymbol{{A}}:\:\frac{\left({x}−\mathrm{2}\right)\left({x}+\mathrm{1}\right)}{{x}+\mathrm{3}}\geqslant\:\mathrm{0} \\ $$$$\:\:\:\:\:\:{x}\:\:\:\:\:\:\:−\infty\:\:\:\:\:\:\:\:\:\:\:−\mathrm{7}\:\:\:\:\:\:\:\:\:\:\cancel{−\mathrm{3}}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:+\infty \\ $$$$\left({x}−\mathrm{2}\right)\left({x}+\mathrm{1}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:−\:\:\:\:\:\:\:\:\:\:\:\:−\:\:\:\:\:\mathrm{0}\:\:\:\:\:+ \\ $$$$\:\:\:\:{x}+\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\:\:\:\:\mathrm{0}\:\:\:\:\:+\:\:\:\:\:\:\:\:\:\:\:\:+ \\ $$$$\:\:\:{Frac}.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\:\:\:\mathrm{0}\:\:\:\:\:\:+\:\:\:\:\:\:\parallel\:\:\:\:\:−\:\:\:\:\mathrm{0}\:\:\:\:\:+ \\ $$$$ \\ $$$$\:{Either}\:\boldsymbol{{x}}\in\:\left[−\mathrm{7},\:−\mathrm{3}\left[\:\cup\:\left[\mathrm{2},\:+\infty\left[\right.\right.\right.\right. \\ $$$$ \\ $$$$\boldsymbol{{B}}:\:\:\frac{\left({x}+\frac{\mathrm{9}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{89}}{\mathrm{4}}}{{x}+\mathrm{3}}\leqslant\:\mathrm{0}\: \\ $$$$\:{x}=\:\frac{−\mathrm{9}−\:\sqrt{\mathrm{89}}}{\mathrm{2}}\:\cong\:−\mathrm{9}.\mathrm{21}\:\:{or}\:{x}=\:\frac{−\mathrm{9}+\:\sqrt{\mathrm{89}}}{\mathrm{2}}\:\cong\:\mathrm{0}.\mathrm{21} \\ $$$$ \\ $$$$\:\:\:\:\:\:{x}\:\:\:\:\:\:\:−\infty\:\:\:\:\:\:\:\:\:\:\:\:\:{x}_{\mathrm{1}} \:\:\:\:\:\:\:\:\:\:\cancel{−\mathrm{3}}\:\:\:\:\:\:\:\:\:\:\:\:{x}_{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:+\infty \\ $$$$\left({x}−\mathrm{2}\right)\left({x}+\mathrm{1}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:−\:\:\:\:\:\:\:\:\:\:\:\:−\:\:\:\:\:\mathrm{0}\:\:\:\:\:+ \\ $$$$\:\:\:\:{x}+\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\:\:\:\:\mathrm{0}\:\:\:\:\:+\:\:\:\:\:\:\:\:\:\:\:\:+ \\ $$$$\:\:\:{Frac}.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\:\:\:\mathrm{0}\:\:\:\:\:\:\cancel{+}\:\:\:\:\:\:\parallel\:\:\:\:\:−\:\:\:\:\mathrm{0}\:\:\:\:\:\cancel{+} \\ $$$$ \\ $$$$\left.\:\left.{O}\left.{r}\left.\:{x}\in\:\right]−\infty,\:\frac{−\mathrm{9}−\:\sqrt{\mathrm{89}}}{\mathrm{2}}\:\right]\:\cup\:\right]−\mathrm{3},\:\frac{−\mathrm{9}+\:\sqrt{\mathrm{89}}}{\mathrm{2}}\:\right] \\ $$$$ \\ $$

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