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if-x-3-lt-2-then-prove-that-x-2-9-16-




Question Number 167296 by mkam last updated on 12/Mar/22
if ∣ x−3∣<2 then prove that ∣x^2 −9∣≤ 16
$$\boldsymbol{{if}}\:\mid\:\boldsymbol{{x}}−\mathrm{3}\mid<\mathrm{2}\:\boldsymbol{{then}}\:\boldsymbol{{prove}}\:\boldsymbol{{that}}\:\mid\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{9}\mid\leqslant\:\mathrm{16} \\ $$
Commented by mkam last updated on 12/Mar/22
??????
$$?????? \\ $$
Commented by cortano1 last updated on 12/Mar/22
 ⇒−2<x−3<2  ⇒−2+6<x−3+6<2+6  ⇒4<x+3<8    { ((−2<x−3<2)),((4 <x+3<8)) :}⇒−16<(x−3)(x+3)<16  ⇒−16<x^2 −9<16  ⇒∣x^2 −9∣<16
$$\:\Rightarrow−\mathrm{2}<\mathrm{x}−\mathrm{3}<\mathrm{2} \\ $$$$\Rightarrow−\mathrm{2}+\mathrm{6}<\mathrm{x}−\mathrm{3}+\mathrm{6}<\mathrm{2}+\mathrm{6} \\ $$$$\Rightarrow\mathrm{4}<\mathrm{x}+\mathrm{3}<\mathrm{8} \\ $$$$\:\begin{cases}{−\mathrm{2}<\mathrm{x}−\mathrm{3}<\mathrm{2}}\\{\mathrm{4}\:<\mathrm{x}+\mathrm{3}<\mathrm{8}}\end{cases}\Rightarrow−\mathrm{16}<\left(\mathrm{x}−\mathrm{3}\right)\left(\mathrm{x}+\mathrm{3}\right)<\mathrm{16} \\ $$$$\Rightarrow−\mathrm{16}<\mathrm{x}^{\mathrm{2}} −\mathrm{9}<\mathrm{16} \\ $$$$\Rightarrow\mid\mathrm{x}^{\mathrm{2}} −\mathrm{9}\mid<\mathrm{16} \\ $$
Commented by MJS_new last updated on 12/Mar/22
I beg your pardon, next time I′ll set my alarm  half an hour earlier so that I can answer your  questions in time.
$$\mathrm{I}\:\mathrm{beg}\:\mathrm{your}\:\mathrm{pardon},\:\mathrm{next}\:\mathrm{time}\:\mathrm{I}'\mathrm{ll}\:\mathrm{set}\:\mathrm{my}\:\mathrm{alarm} \\ $$$$\mathrm{half}\:\mathrm{an}\:\mathrm{hour}\:\mathrm{earlier}\:\mathrm{so}\:\mathrm{that}\:\mathrm{I}\:\mathrm{can}\:\mathrm{answer}\:\mathrm{your} \\ $$$$\mathrm{questions}\:\mathrm{in}\:\mathrm{time}. \\ $$
Answered by MJS_new last updated on 12/Mar/22
∣x−3∣<2  ⇒ 1<x<5  ⇒ 1<x^2 <25  ⇒ −8<x^2 −9<16  ⇒ ∣x^2 −9∣<16
$$\mid{x}−\mathrm{3}\mid<\mathrm{2} \\ $$$$\Rightarrow\:\mathrm{1}<{x}<\mathrm{5} \\ $$$$\Rightarrow\:\mathrm{1}<{x}^{\mathrm{2}} <\mathrm{25} \\ $$$$\Rightarrow\:−\mathrm{8}<{x}^{\mathrm{2}} −\mathrm{9}<\mathrm{16} \\ $$$$\Rightarrow\:\mid{x}^{\mathrm{2}} −\mathrm{9}\mid<\mathrm{16} \\ $$

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