Question

.....





Express the following set both in Roster and Rule form.......

Please answer...▪️▪️▪️

Don't dare to give Spam answers...❌​

Answer 1

Answer:

Step-by-step explanation:

x³-2x²-x+2=0

x²(x-2)-(x-2)=0

(x-2)(x²-1)=0

(x-2)(x-1)(x+1)=0

x=2,1 and -1

Thus positive roots are (2,1)

So the set S={ 1,2}

S= { x : x is a positive and real root of x³-2x²-x+2=0}

Answer 2

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Question\;:-}}}$

Here the Concept of Sets has been used. We are given the equation. First we can find its roots. The we can sort out the real roots and make its set. After finding the set, we can express it in set builder and roster form.

Let's do it !!

_______________________________________________

★ Solution :-

Given,

» x³ - 2x² - x + 2 = 0

Firstly let's find the roots of this equation. Then,

$\\\;\;\sf{:\rightarrow\;\;x^{3}\;-\;2x^{2}\;-\;x\;+\;2\;=\;\bf{0}}$

Taking the common terms, we get,

$\\\;\;\sf{:\rightarrow\;\;x^{2}(x\;-\;2)\;-\;1(x\;-\;2)\;=\;\bf{0}}$

$\\\;\;\sf{:\rightarrow\;\;(x^{2}\;-\;1)(x\;-\;2)\;=\;\bf{0}}$

Here using the identity,

✒ a² - b² = (a + b)(a - b), we get,

$\\\;\;\sf{:\rightarrow\;\;(x^{2}\;-\;1^{2})(x\;-\;2)\;=\;\bf{0}}$

$\\\;\;\sf{:\rightarrow\;\;(x\;-\;1)(x\;+\;1)(x\;-\;2)\;=\;\bf{0}}$

Since, all are the multiples here. Thus,

Either, (x - 1) = 0, (x + 1) = 0 or (x - 2) = 0

From this, we get,

$\\\;\;\sf{:\Longrightarrow\;\;(x\;-\;1)\;=\;0\;,\;(x\;+\;1)\;=\;0\;,\;(x\;-\;2)\;=\;0}$

$\\\;\;\sf{:\Longrightarrow\;\;x\;=\;1\;,\;\;x\;=\;-1\;,\;\;x\;=\;2}$

Hence, we got, x = 2, 1 and -1

Its given in the question that we need to take out the set of all positive real roots. Thus here we have to neglect -1 since its negative root.

Then,

• x = 2, 1

Now let us take out the sets. Let this set be S. Then,

$\\\;\underline{\boxed{\tt{Hence\;\;set\;\;of\;\;roots,\;S\;\;=\;\;\bf{\{1\;,\;2\}}}}}$

~ Roster Form ::

$\\\;\;\underline{\boxed{\tt{S_{(roster)}\;=\;\bf{\{1\;,\;2\}}}}}$

~ Set - Builder Form ::

$\\\;\;\underline{\boxed{\tt{S_{(set\;-\;builder)}\;=\;\bf{\{x\;:\;x\;=\;n,\;where\;n\;is\;a\;natural\;number\;and\;1\;\leq\;n\;\leq\;2\}}}}}$

_______________________________________________

★ More to know :-

Let a, b ϵ R and a < b.

Then,

$\\\;\sf{\leadsto\;\;[a\;,\;b]\;=\;\{x\;:\;a\;\leq\;x\;\leq\;b\}}$

$\\\;\sf{\leadsto\;\;(a\;,\;b)\;=\;\{y\;:\;a\;<\;y\;<\;b\}}$

$\\\;\sf{\leadsto\;\;[a\;,\;b)\;=\;\{x\;:\;a\;\leq\;x\;<\;b\}}$

$\\\;\sf{\leadsto\;\;(a\;,\;b]\;=\;\{x\;:\;a\;<\;x\;\leq\;b\}}$

$\\\;\sf{\leadsto\;\;Empty\;Set\;=\;\phi\;\;or\;\; \{\}}$