If, $| x + 5 | < \frac{1}{2},$ $|x+5|<1/2,$ then show that $∣ ∣ x^{2} - 25 ∣ ∣ < \frac{21}{4}$ $|x^2-25|<21/4$ Could someone help me to solve the problem?

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Answer 1 · 2018-03-06T17:46:30

Kindly refer to the Explanation.

$Prerequisites : (1) : | a + b | \leq | a | + | b |, (2) : | a b | = | a | | b |$ $"Prerequisites : "(1) : |a+b| le |a|+|b|, (2) : |ab|=|a||b|$ .

$By (2), ∣ ∣ x^{2} - 25 ∣ ∣ = | (x + 5) (x - 5) | = | x + 5 | | x - 5 |$ $"By (2), "|x^2-25|=|(x+5)(x-5)|=|x+5||x-5|$ .

$"Now, "|x+5| lt 1/2............[Given]$ .

Multiplying by $|x-5| gt 0$ , we get,

$|x+5||x-5| lt 1/2|x-5|............(ast1)$ .

But, $|x-5|=|(x+5)+(-10)|$ ,

$le |x+5|+|(-10)|...[because, (1)]$ ,

$=|x+5|+10 lt 1/2+10........[because," Given]"$ .

$rArr |x-5| le 21/2............(ast2)$ .

$(ast1) and (ast2) rArr |x+5||x-5| lt 1/2*21/2, or,$

$|x^2-25| lt 21/4$ , as desired!

If, |x+5|<1/2,|x+5|<12,|x+5|<1/2, then show that|x^2-25|<21/4∣∣x2−25∣∣<214|x^2-25|<21/4 Could someone help me to solve the problem?