Surds & Length of straight line

[synkk]

Homework Statement

1)The normal to C ( 3y = x + 20 ) at P (4, 8) cuts the x-axis at the point Q.
Find the length PQ, giving your answers in a simplified surd form.

2) write \dfrac{2\sqrt{x} + 3}{x} in the form 2x^p + 3x^q where p and q are constants

Homework Equations

y = mx + c? I'm not sure

The Attempt at a Solution

for 1) i let y = 0 and got Q (-20,0) but i don't know how to find the length of the two points.

for 2)
\dfrac{3}{x} = 3x^{-1} but i don't know what \dfrac{2\sqrt{x}}{x} is.

thanks

 

[LCKurtz]

Homework Statement

1)The normal to C ( 3y = x + 20 ) at P (4, 8) cuts the x-axis at the point Q.
Find the length PQ, giving your answers in a simplified surd form.

The Attempt at a Solution

for 1) i let y = 0 and got Q (-20,0) but i don't know how to find the length of the two points.

Q is the point where the normal intersects the x axis. You have to start by writing the equation of the normal line. And you don't find the "length of two points". You find the distance between the two points.

 

[LCKurtz]

2) write \dfrac{2\sqrt{x} + 3}{x} in the form 2x^p + 3x^q where p and q are constants


\dfrac{3}{x} = 3x^{-1} but i don't know what \dfrac{2\sqrt{x}}{x} is.

thanks

Write it with exponents and use the rules of exponents to simplify it.

 

[synkk]

Write it with exponents and use the rules of exponents to simplify it.

Alright, for 1)

I let y = 0 and got Q as (-20,0) (seeing as it says it crosses the x-axis)

so <br /> \sqrt{(4-(-20))^2 + (8-0)^2}<br /> = \sqrt{640}<br /> = 8\sqrt{10}<br />
for 2)

\dfrac{2\sqrt{x}}{x} = 2x^{-\dfrac{1}{2}}<br /> <br /> = 2x^{-\dfrac{1}{2}} + 3x^{-1}<br /> <br />

Have i gone wrong anywhere?

 

[LCKurtz]

Alright, for 1)

I let y = 0 and got Q as (-20,0) (seeing as it says it crosses the x-axis)

Did you even read my post?? Q = (-20,0) has nothing to do with the problem.

for 2)

\dfrac{2\sqrt{x}}{x} = 2x^{-\dfrac{1}{2}}<br /> <br /> = 2x^{-\dfrac{1}{2}} + 3x^{-1}<br /> <br />

You have worked the exponents correctly, but that last line has = signs between things that aren't equal.

 

[synkk]

Did you even read my post?? Q = (-20,0) has nothing to do with the problem.


You have worked the exponents correctly, but that last line has = signs between things that aren't equal.

I wrote = just to show it goes to that answer, but whatever i understand that question now, thanks.

For 1) the question says that the normal to C at P is 3y = x + 20 (its on the question before that), P is (4,8) also on the question before that.

It says that the normal at (4,8) cuts the x-axis at Q, to find where it crosses the x-axis you would have to let y = 0 no? If not I'm completely mistaken as that's the only way I'm seeing the question at the moment, i don't see what your trying to say.

thanks again.

 

[LCKurtz]

Homework Statement

1)The normal to C ( 3y = x + 20 ) at P (4, 8) cuts the x-axis at the point Q.

For 1) the question says that the normal to C at P is 3y = x + 20 (its on the question before that), P is (4,8) also on the question before that.

We aren't mind readers here. The statement I have bolded is readily interpreted "The normal to [the curve] C [whose equation is] 3y = x + 20 ... meaning you have this curve C whose equation is given and the first problem is to find its normal. Why do you even mention it is normal to some curve C which isn't given and is irrelevant? How are we supposed to know what is on "the question before that"?

 

[synkk]

We aren't mind readers here. The statement I have bolded is readily interpreted "The normal to [the curve] C [whose equation is] 3y = x + 20 ... meaning you have this curve C whose equation is given and the first problem is to find its normal. Why do you even mention it is normal to some curve C which isn't given and is irrelevant? How are we supposed to know what is on "the question before that"?

I thought it was pretty clear that the NORMAL TO C (3y=x+20), i didn't want to mention other questions because I've already solved the other questions and mentioned the data which was needed for this question...

thanks for your help anyway, appreciated.