Question

. Let A = {x: x ϵ R, x2 – 5x + 6 = 0 } and B = { x: x ϵ R, x2 = 9}. Find A intersection B and A union B

Answer 1

SETS :

For A = {x: x ∊ R, x^2 – 5x + 6 = 0 }

A = { x^2 - 2x - 3x + 6 = 0}

A = { x( x - 2) - 3 ( x - 2) = 0}

∴ A = { 2,3}

B = { x^2 = 9}

∴ B = { 3}

⛶ A ∪ B = { 2,3}

⛶ A ∩ B = { 3 }

⛛ Union of A & B set = All elements of A & B.

⛛ Intersection of A & B set = Common elements in A & B.

Answer 2

Answer:

The A intersection B is the Roster form of the set A = { 2,3}

Roster form of the set B = { -3,3}

AUB = {-3,2,3}

AnB = {3}

Step-by-step explanation:

• Sets, in mathematics, exist as a systematized array of objects and can be described in set-builder form or roster form.
• Set-builder notation is a mathematical inscription for representing a set by representing its components or describing the properties that its members must satisfy.
• In roster form, all the elements of a set exist listed, the elements are being divided by commas and are enclosed within braces { }

Given:

Let A = {x: x ϵ R, $x^{2}$– 5x + 6 = 0 }

B = { x: x ϵ R, $x^{2}$ = 9}

To find:

AUB and AnB

Step 1

Given,

$x^{2} -5x+6 = 0$

⇒$x^{2} -2x-3x+6 = 0$

⇒$x(x-2)-3(x-2) = 0$

⇒$(x-2)(x-3) = 0$

Therefore,

$x-2 = 0$

and $x-3 = 0$

The roots of x²-5x+6 = 0 exist 2 and 3

Roster forms of A

A = { 2,3}

and

B={x:x belongs to R, x²=9}

⇒ x² = 9

⇒x =±√9

⇒x = ±3

⇒ x = 3 or -3

Roots of x²=9 exist 3 and -3

Roster form of B

B ={-3,3}

Then,

AUB

⇒ {2,3} U { -3,3}

⇒AUB = {-3,2,3}

AnB = {2,3} n { -3,3}

⇒ AnB = {3}.

Therefore,

Roster form of the set A = { 2,3}

Roster form of the set B = { -3,3}

AUB = {-3,2,3}

AnB = {3}

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