Question

1)Identify the conjugate for 6 – 7√3 and explain your reasoning.

Answer 1

Answer:

1) The conjugate for $6-7\cdot \sqrt{3}$ is $6+7\cdot \sqrt{3}$.

2) -111

3) It is because number and its conjugate have the same numbers but their second components are opposite to each other.

Step-by-step explanation:

1) A radical number of the form $r = a + b\sqrt{c}$, $\forall \,a,b,c\in \mathbb{R}$, where the first component is a non-radical real number which acts as a "pivot" and the second component is the radical component, which generates a "displacement" from pivot.

The conjugate of $r$ is a real number $s = a - b\sqrt{c}$,  $\forall \,a,b,c\in \mathbb{R}$, which that is "pivot" plus "displacement" in the direction opposite to that of $r$. We proceed to complement this explanation with the image attached below.

Then, the conjugate for $6-7\cdot \sqrt{3}$ is $6+7\cdot \sqrt{3}$.

2) We proceed to performed all the need algebraic operation until result is found:

1) $(6+7\sqrt{3})\cdot (6-7\sqrt{3})$ Given

2) $6\cdot (6-7\sqrt{3})+7\sqrt{3}\cdot (6-7\sqrt{3})$ Distributive property

3) $6\cdot [6+(-7)\cdot \sqrt{3}]+7\cdot \sqrt{3}\cdot [6+(-7)\cdot \sqrt{3}]$ Definition of subtraction/ $-a\cdot b = (-a)\cdot b$

4) $6\cdot 6 +[6\cdot (-7)]\cdot \sqrt{3}+(6\cdot 7)\cdot \sqrt{3}+[7\cdot (-7)]\cdot (\sqrt{3}\cdot \sqrt{3})$ Distributive, associative and commutative properties

5) $36 +(-42)\cdot \sqrt{3}+42\cdot \sqrt{3}+(-49)\cdot 3$ Definition of multiplication/$-a\cdot b = (-a)\cdot b$/Definition of square root

6) $[36 +(-147)]+\sqrt{3}\cdot [42+(-42)]$ Commutative, associative and distributive properties/$-a\cdot b = (-a)\cdot b$

7) $-111+\sqrt{3}\cdot 0$ Definition of subtraction/Existence of additive inverse

8) $-111+0$ $a\cdot 0 = 0$

9) $-111$ Modulative property/Result

3) It is because number and its conjugate have the same numbers but their second components are opposite to each other.