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1)The range of values of "$a$", such that

$|x-2|< a$ is a necessary condition for $x^2-3x-10<0$

2)The range of values of "$a$", such that

$|x-2| < a$ is a sufficient condition for $x^2-3x-10<0$

I have found that a necessary and sufficient condition for $x^2-3x-10<0$ is $-2< x <5$
, how can I answer that two problems? and what is the different between necessary, sufficient, necessarry and sufficient condition? is there about logical mathematics ?

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  • $\begingroup$ $2<x<1$ is satisfied by no real number. Also how did you get from one to the other? It seems to me that, for example, 3 satisfies the inequality. $\endgroup$ – Niki Di Giano Dec 18 '18 at 0:50
  • $\begingroup$ @NikiDiGiano, sorry I made a mistake, I have correct them, -2<x<5 $\endgroup$ – Aster Zen Dec 18 '18 at 0:59
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Necessary, but not sufficient condition: $$a>0,$$ because the absolute value inequality must hava a silution.

Necessary and sufficient condition: $$x^2-3x-10<0 \Rightarrow -2<x<5 \Rightarrow -4<x-2<3 \Rightarrow |x-2|<3 \Rightarrow a\le 3;\\ 0<a\le 3.$$

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  • $\begingroup$ how can u get the -4<x<3, do you subtract the -2<x<5 with 2 ?, and how can you get the |x−2|<4⇒a≥4 ? $\endgroup$ – Aster Zen Dec 18 '18 at 1:11
  • $\begingroup$ Yes, sorry, you minus $2$ from all three sides. $\endgroup$ – farruhota Dec 18 '18 at 1:15
  • $\begingroup$ Because, $|x-2|<4\le a$. $\endgroup$ – farruhota Dec 18 '18 at 1:17
  • $\begingroup$ hmm sorry, Can you explain more on how can u get this : −4<x−2<3⇒|x−2|<4⇒a≥4, I still confused, $\endgroup$ – Aster Zen Dec 18 '18 at 1:21
  • $\begingroup$ Sorry I’ve corrected: $0<a\le 3$. For example, when $a=3$, $|x-2|<3 \Rightarrow \\ -1<x<5 \Rightarrow \\ -2<x<5 \Rightarrow \\ x^2-3x-10<0.$ $\endgroup$ – farruhota Dec 18 '18 at 1:57
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Necessary is what it absolutely must have.

To be a dessert it is necessary that it be food.

To show if $|x-2| < a$ is necessary condition for $x^2 - 3x -10 <$ you must prove

$x^2 - 3x -10 > 0 \implies |x-2| < a$.

Sufficient is something it doesn't need to be, but if it is that's enough to show it is true.

To be a dessert it is sufficient to be ice cream.

(Not all desserts are ice cream, but all ice creams are desserts. So if you can prove something is an ice cream that is enough, or in other words, that is sufficient to prove it is a dessert.

To show if $|x-2| < a$ is necessary condition for $x^2 - 3x -10 <$ you must prove

$|x-2| < a \implies x^2 - 3x -10 > 0$.

Nescessary and sufficient is both. It be needs to be and if it is that will prove it.

To be a dessert is is necessary and sufficient to be a sweet food eaten after dinner.

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  • $\begingroup$ and how can I prove that ? $\endgroup$ – Aster Zen Dec 18 '18 at 1:52
  • $\begingroup$ Solve. $x^2 - 3x - 10 = (x-5)(x + 2) < 0$ implies. This will happen will happen if and only if one of $x-5$ and $x+2$ is greater than $0$ and then other is less. As $x+2 > x-5$ this will only happen if $x+2 > 0 > x-5$ if and only if $-2 < x < 5$ if and only if $-4 < x-2 < 3$. This can't be true if $|x-2| \ge 4$ so for any $a\ge 4$ it is nescessary that $|x-2|< a$. This will always be true if $|x-2| < 3$ so for any $a < 3$ it is sufficient that that $|x-2| < a$. But there is no $a$ for which it'd be nesc. and sufficient. $\endgroup$ – fleablood Dec 18 '18 at 6:44
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Saying that $A$ is necessary for $B$ means that $B$ implies $A$-that any time $B$ is true, so is $A$. If $B$ is false, $A$ may still be true.

Saying that $A$ is sufficient for $B$ means that $A$ implies $B$.

Saying that $A$ is necessary and sufficient for $B$ means both of the above are true, so $A$ and $B$ are either both true or both false.

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  • $\begingroup$ but how can I apply that rules on the question above? It include absolute number $\endgroup$ – Aster Zen Dec 18 '18 at 1:00
  • $\begingroup$ $a$ is how far $x$ is from $2$. You know the range of $x$. Sufficient means that if $x$ is within $a$ of $2$ then the polynomial is negative. Any time $|x-2| \lt 3 we have $-1 \lt x \lt 5$, so the polynomial will be negative. $\endgroup$ – Ross Millikan Dec 18 '18 at 1:21
  • $\begingroup$ oh I realized that the distance from -2 to 2 is 4, and the distance from 5 to 2 is 3, So is it about how far the range of x from 2 ? is it just as simple as that ? $\endgroup$ – Aster Zen Dec 18 '18 at 1:27
  • $\begingroup$ Yes, that is one useful way to think of the absolute value. $\endgroup$ – Ross Millikan Dec 18 '18 at 2:48

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