Methods Unit 1&2 q's

[idontknow2298]

Find the equation of the diameter of the circle x^2+y^2-4x+6y=14

[Phenomenol]

Find the equation of the diameter of the circle x^2+y^2-4x+6y=14

Ughhhh, do you mean the equation of the circle or just the diameter of the circle?

[alchemy]

So I've got a question but it doesn't seem doable.
Let m and n be the roots of the quadratic equation 4x^2+5x+3=0. Find (m+7)(n+7).
Firstly, we can see that it has no solutions, by finding the discriminant and graphing! Then, I tried using Vieta's formula to find the sum and product of m and n, but that wouldn't work since it has no solutions...
I wouldn't be surprised if this was a book question that had an error, but this is coming from a past test. Am I doing something wrong?

[Phenomenol]

So I've got a question but it doesn't seem doable.
Let m and n be the roots of the quadratic equation 4x^2+5x+3=0. Find (m+7)(n+7).
Firstly, we can see that it has no solutions, by finding the discriminant and graphing! Then, I tried using Vieta's formula to find the sum and product of m and n, but that wouldn't work since it has no solutions...
I wouldn't be surprised if this was a book question that had an error, but this is coming from a past test. Am I doing something wrong?

You don't seem to be doing anything wrong. This question could be answered if you took complex roots, though.

[K3NUpdate]

Find the equation of the diameter of the circle x^2+y^2-4x+6y=14

 idontknow2298, did you mean: Find the equation of the equation of the diameter of the circle x^2+y^2-4x+6y=14 which passes through the origin?

If this is the case, then the way to do this is to first complete the square to give you the equation of the circle, which should give you:
(x-2)^2+(x+3)^2=27

The centre is (2,-3) and the radius is 3√3.

Then, you find the gradient by using the two points lying on this line, which are (2,-3) and (0,0), since the diameter passes through the origin.

M=0-2/0-(-3)
M=-2/3

Sub the gradient into y=mx+c.

y=-2/3x+c

Sub in either (2,-3) or (0,0) into y=-2/3x+c to find the y-intercept.

Sub (0,0)
c=0

Therefore, we get the equation of the diameter of the circle:

y=-2/3x

You can also put this into ax+by+c=0, if it's necessary.

[knightrider]

how do you draw a function when there is an unknown factor?

For example f(x)=k(x-2)^2+3

[rhinwarr]

how do you draw a function when there is an unknown factor?

For example f(x)=k(x-2)^2+3

Same as normal except some points will be known, some will have an unknown.
eg. The turning point will be (2,-3), the y intercept will be (0,4k+3)

[knightrider]

Same as normal except some points will be known, some will have an unknown.
eg. The turning point will be (2,-3), the y intercept will be (0,4k+3)

thx but how would you draw the yintecept on the graph if it is 4k+3 and how would you do it using a cas calculator

[rhinwarr]

I don't think you can do it with a CAS calculator.
For hand drawing, put in the points that you know, then join them with the appropriately shaped graph. Then you can just label the co-ordinates of the intercept, wherever it may be on the graph. The main thing with drawing these graphs is the shape and if the unknown has a limited set of numbers, which side of the axes the graph is on.

[knightrider]

I don't think you can do it with a CAS calculator.
For hand drawing, put in the points that you know, then join them with the appropriately shaped graph. Then you can just label the co-ordinates of the intercept, wherever it may be on the graph. The main thing with drawing these graphs is the shape and if the unknown has a limited set of numbers, which side of the axes the graph is on.

which side of the axes is the unknown on in this case can you please draw it for me i mean the whole graph that would really eb helpful and thx so much fro your help

[idontknow2298]

How would I do this question?
Without dividing, show that the first polynomial is exactly divisible by the second polynomial:
2x^3-13x^2+27x-18, 2x-3

[IndefatigableLover]

How would I do this question?
Without dividing, show that the first polynomial is exactly divisible by the second polynomial:
2x^3-13x^2+27x-18, 2x-3

We can assume that 2x-3 is a factor of the first polynomial.

If we said 2x-3=0, then we know that x=3/2

So normally when we're testing for factors, we let P(x)=_____________________

In this case we already know our factor so we just do the same thing and input it in!

[idontknow2298]

thank you so much