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x_I-2+y=3+_I-18  

_I- is root

 

What is x and y? How you found them?

 Aug 15, 2014

Best Answer 

 #1
avatar+118723 
+10

There are a couple of different ways of inserting a square root sign

1)    Under insert there are special characters and √ is on the 3rd row in the middle.

OR

2)   you can use the math formula button

 x*sqrt(2)+y=3+sqrt(18)  would look like 

 

$${\mathtt{x}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{y}} = {\mathtt{3}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{18}}}}$$

 

3)   I use the LaTex tab but the 2nd option is probably the best for you.

 

$$\\x\sqrt2+y=3+\sqrt{18}\\
y=-x\sqrt2 +3+\sqrt{18}\\
y=-\sqrt2 x +3+\sqrt{9*2}\\
y=-\sqrt2 x +3(1+\sqrt{2})\\$$

 

Now this is the equation of a straight line (linear equation) so there are an infinite number of solutions.

 

The gradient is $${\mathtt{\,-\,}}{\sqrt{{\mathtt{2}}}}$$   and the y intercept is  $${\mathtt{3}}{\mathtt{\,\times\,}}\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{2}}}}\right)$$

 Aug 15, 2014
 #1
avatar+118723 
+10
Best Answer

There are a couple of different ways of inserting a square root sign

1)    Under insert there are special characters and √ is on the 3rd row in the middle.

OR

2)   you can use the math formula button

 x*sqrt(2)+y=3+sqrt(18)  would look like 

 

$${\mathtt{x}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{y}} = {\mathtt{3}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{18}}}}$$

 

3)   I use the LaTex tab but the 2nd option is probably the best for you.

 

$$\\x\sqrt2+y=3+\sqrt{18}\\
y=-x\sqrt2 +3+\sqrt{18}\\
y=-\sqrt2 x +3+\sqrt{9*2}\\
y=-\sqrt2 x +3(1+\sqrt{2})\\$$

 

Now this is the equation of a straight line (linear equation) so there are an infinite number of solutions.

 

The gradient is $${\mathtt{\,-\,}}{\sqrt{{\mathtt{2}}}}$$   and the y intercept is  $${\mathtt{3}}{\mathtt{\,\times\,}}\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{2}}}}\right)$$

Melody Aug 15, 2014

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