MHB Double check my expressions a+b squareroot c

[ai93]

Can anyone double check my answers? I have to put the answer in this format $$a+b\sqrt{c}$$
a)$$(3-\sqrt{18})^{2}$$
Answer
$$(3-\sqrt{18}) \quad(3-\sqrt{18})$$
$$9-3\sqrt{18}-3\sqrt{18}+18$$
$$9-6\sqrt{18}+18$$
$$27(-6)\quad(3\sqrt{2)}$$
=$$27-18\sqrt{2}$$
$$\therefore a = 27 b = -18 c = \sqrt{2}$$
Wouldn't I have to put it in $$+b?$$ I got -18

b) $$\frac{3}{4-\sqrt{7}}$$
Answer
$$\frac{3}{4-\sqrt{7}}$$ x $$\frac{4+\sqrt{7}}{4+\sqrt{7}}$$
$$\frac{3(4+\sqrt{7)}}{(4-\sqrt{7)(4+\sqrt{7)}}}$$
$$\frac{12+3\sqrt{7}}{16+4\sqrt{7}-4\sqrt{7}-7}$$
So
$$\frac{12+3\sqrt{7}}{9}$$ CF is 3 so divide by 3
=$$\frac{4}{3}+1\sqrt{7}$$
$$a=\frac{4}{3} b=1 c=\sqrt{2}$$

c) $$\frac{4+\sqrt{5}}{3-\sqrt{5}}$$
Answer
$$\frac{4+\sqrt{5}}{3-\sqrt{5}}$$ x $$\frac{3+\sqrt{5}}{3+\sqrt{5}}$$
$$\frac{(4+\sqrt{5)(3+\sqrt{5)}}}{(3-\sqrt{5)(3+\sqrt{5)}}}$$
$$\frac{12+4\sqrt{5}+3\sqrt{5}+5}{9+3\sqrt{5-3\sqrt{5}-5}}$$
=$$\frac{17}{4}+7\sqrt{5}$$
$$a=\frac{17}{4} b= 7 c=\sqrt{5}$$

Please correct me if i am wrong, and some of the questions I cannot display how I want, I hope you understand

 

[ineedhelpnow]

no you wouldn't change it to +b. the simple form of the expression is a+b. instead of having both forms $a+b \sqrt{c}$ and $a-b \sqrt{c}$. they just have a+b which also implies $a+(-b) \sqrt{c}$ where b is negative which can be $a-b \sqrt{c}$ in the final form.

 

[Evgeny.Makarov]

a)$$(3-\sqrt{18})^{2}$$
Answer
$$(3-\sqrt{18}) (3-\sqrt{18})$$
$$9-3\sqrt{18}-3\sqrt{18}+18$$
$$9-6\sqrt{18}+18$$
$$27(-6)(3\sqrt{2)}$$
=$$27-18\sqrt{2}$$
$$\therefore a = 27 b = 18 c = \sqrt{2}$$

This is correct. (Edit: Not entirely correct because $b=-18$.) (Hint: Use [m]\ [/m] (backslash space) or [m]\quad[/m] in LaTeX to insert a space in math mode. Or, better, use several [m][math][/m] tags.)

Wouldn't I have to put it in $$+b?$$ I got -18

You've got $+b$ where $b=-18$.

b) $$\frac{3}{4-\sqrt{7}}$$
Answer
$$\frac{3}{4-\sqrt{7}}$$ x $$\frac{4+\sqrt{7}}{4+\sqrt{7}}$$
$$\frac{3(4+\sqrt{7)}}{(4-\sqrt{7)(4+\sqrt{7)}}}$$

Close the curly brace after 7 in the denominator, like this: [m]\sqrt{7}[/m] instead of this: [m]\sqrt{7)[/m].

$$\frac{12+3\sqrt{7}}{16+4\sqrt{7}-4\sqrt{7}-7}$$
So
$$\frac{12+3\sqrt{7}}{9}$$ CF is 3 so divide by 3
=$$\frac{4}{3}+1\sqrt{7}$$

Why did you divide 12 by 9. but did not divide 3 by 9?

c) $$\frac{4+\sqrt{5}}{3-\sqrt{5}}$$
Answer
$$\frac{4+\sqrt{5}}{3-\sqrt{5}}$$ x $$\frac{3+\sqrt{5}}{3+\sqrt{5}}$$
$$\frac{(4+\sqrt{5)(3+\sqrt{5)}}}{(3-\sqrt{5)(3+\sqrt{5)}}}$$

Same remark about closing }.

$$\frac{12+4\sqrt{5}+3\sqrt{5}+5}{9+3\sqrt{5-3\sqrt{5}-5}}$$
=$$\frac{17}{4}+7\sqrt{5}$$

Why did you divide 17 by 4, but did not divide 7 by 4?

 

[ai93]

This is correct. (Hint: Use [m]\ [/m] (backslash space) or [m]\quad[/m] in LaTeX to insert a space in math mode. Or, better, use several [m][math][/m] tags.)

You've got $+b$ where $b=-18$.

Close the curly brace after 7 in the denominator, like this: [m]\sqrt{7}[/m] instead of this: [m]\sqrt{7)[/m].

Why did you divide 12 by 9. but did not divide 3 by 9?

Same remark about closing }.

Why did you divide 17 by 4, but did not divide 7 by 4?

Made the correction. a) is equal to $$a = 27 b = -18 c = \sqrt{2}$$

When I try and edit the curly brace \sqrt{7} instead of \sqrt{7) I get this text:\displaystyle \frac{3(4+\sqrt{7)}}{(4-\sqrt{7}(4+\sqrt{7)}}}

I don't quite understand. I divided 12 by 9 because in my step earlier I had $$\frac{12+3\sqrt{7}}{16+4\sqrt{7}-4\sqrt{7}-7}$$
$$+4\sqrt{7}-4\sqrt{7}$$ Both cancel each other out, leaving 16-7=9
I thought you would have to show the answer like this $$\frac{12}{9}+3\sqrt{7}$$ Use a CF to simplify which is 3 and got my answer $$\frac{4}{3}+1\sqrt{7}$$ ?

Same with what I thought with c) Cancel the like terms $$3\sqrt{5}-3\sqrt{5}$$ leaving $$9-5=4$$

I thought you only divide the denominator only by the first part and not necessary to divide by the whole numerator?

 

[Evgeny.Makarov]

Made the correction. a) is equal to $$a = 27 b = -18 c = \sqrt{2}$$

The value of $c$ is 2 since the answer should be in the form $a+b\sqrt{c}$.

When I try and edit the curly brace \sqrt{7} instead of \sqrt{7) I get this text:\displaystyle \frac{3(4+\sqrt{7)}}{(4-\sqrt{7}(4+\sqrt{7)}}}

You still have \sqrt{7) in two places. The correct expression should be [m]\displaystyle \frac{3(4+\sqrt{7})}{(4-\sqrt{7})(4+\sqrt{7})}[/m], which gives
\[
\frac{3(4+\sqrt{7})}{(4-\sqrt{7})(4+\sqrt{7})}.
\]

I thought you only divide the denominator only by the first part and not necessary to divide by the whole numerator?

When you simplify $(12+3)\cdot9$, you multiply both terms inside parentheses by 9:
\[
(12+3)\cdot9=12\cdot9+3\cdot 9
\]
You don't multiply just 12:
\[
(12+3)\cdot9\ne12\cdot9+3
\]
The same happens with division:
\[
(12+3)\cdot\frac{1}{9}=12\cdot\frac{1}{9}+3\cdot\frac{1}{9}.
\]
And the same happens with your expression:
\[
(12+3\sqrt{7})\cdot\frac{1}{9}=12\cdot\frac{1}{9}+3\sqrt{7}\cdot\frac{1}{9}.
\]

 

[ai93]

The value of $c$ is 2 since the answer should be in the form $a+b\sqrt{c}$.

You still have \sqrt{7) in two places. The correct expression should be [m]\displaystyle \frac{3(4+\sqrt{7})}{(4-\sqrt{7})(4+\sqrt{7})}[/m], which gives
\[
\frac{3(4+\sqrt{7})}{(4-\sqrt{7})(4+\sqrt{7})}.
\]

When you simplify $(12+3)\cdot9$, you multiply both terms inside parentheses by 9:
\[
(12+3)\cdot9=12\cdot9+3\cdot 9
\]
You don't multiply just 12:
\[
(12+3)\cdot9\ne12\cdot9+3
\]
The same happens with division:
\[
(12+3)\cdot\frac{1}{9}=12\cdot\frac{1}{9}+3\cdot\frac{1}{9}.
\]
And the same happens with your expression:
\[
(12+3\sqrt{7})\cdot\frac{1}{9}=12\cdot\frac{1}{9}+3\sqrt{7}\cdot\frac{1}{9}.
\]

I am slightly confused. So in regards to my questions, have I got them right so far until the simplifying part? So what I would have to do is $$\frac{1}{9}\cdot\quad(12+3\sqrt{7})$$ .. onto each individual part as you have explained. $$12\cdot\frac{1}{9}$$ equal to $$\frac{4}{3}$$? How would you times $$3\sqrt{7}\cdot\frac{1}{9}$$?

 

[Evgeny.Makarov]

So in regards to my questions, have I got them right so far until the simplifying part?

I believe so, though this depends on what you mean by the "simplifying part". It could be argued that the whole transformation you are doing is simplification.

So what I would have to do is $$\frac{1}{9}\cdot\quad(12+3\sqrt{7})$$ .. onto each individual part as you have explained.

Yes.

$$12\cdot\frac{1}{9}$$ equal to $$\frac{4}{3}$$?

Yes.

How would you times $$3\sqrt{7}\cdot\frac{1}{9}$$?

Well, $3\cdot\sqrt{7}\cdot\frac{1}{9}=3\cdot\frac19\cdot\sqrt7$. I hope you can simplify $3\cdot\frac19$?

All in all,
\[
\frac{a+b}{c}=(a+b)\cdot\frac{1}{c}=a\cdot\frac{1}{c}+b\cdot\frac{1}{c}=\frac{a}{c}+\frac{b}{c}.
\]
Knowing this law is a prerequisite to trasformations you are trying to do.

 

[ai93]

I believe so, though this depends on what you mean by the "simplifying part". It could be argued that the whole transformation you are doing is simplification.

Yes.

Yes.

Well, $3\cdot\sqrt{7}\cdot\frac{1}{9}=3\cdot\frac19\cdot\sqrt7$. I hope you can simplify $3\cdot\frac19$?

All in all,
\[
\frac{a+b}{c}=(a+b)\cdot\frac{1}{c}=a\cdot\frac{1}{c}+b\cdot\frac{1}{c}=\frac{a}{c}+\frac{b}{c}.
\]
Knowing this law is a prerequisite to trasformations you are trying to do.

$$3\cdot\frac{1}{9}$$ equal to $$\frac{1}{3}$$? My only doubts would be when you times say $$\frac{1}{3}$$ by a square root e.g $$\sqrt{7}$$

 

[Evgeny.Makarov]

My only doubts would be when you times say $$\frac{1}{3}$$ by a square root e.g $$\sqrt{7}$$

You can multiply both sides of an equation by the same number, e.g.:
\[
3\cdot\frac{1}{9}=\frac{1}{3}\implies
3\cdot\frac{1}{9}\cdot\sqrt{7}=\frac{1}{3}\cdot\sqrt{7}.
\]