questions (1 Viewer)

[Arithela]

What do I do when asked to find the reciprocals roots of a quadratic?


2. A straight line, L, passing through (2,0) meets the parabola y = x^2 in A and B.

Use the point gradient formula to show that the equation of the line L is y = mx - 2m, where m is its gradient.

 

[Arithela]

Thanks!

This is a follow-on from question 2.

(ii) Use simultaneous equations to show that the x-coordinates of A and B are given by the roots of the equation x^2 - mx + 2m = 0

(iii) find the co-ordinates of m, the midpoint of AB

 

[Arithela]

Thanks!

This is a follow-on from question 2.

(ii) Use simultaneous equations to show that the x-coordinates of A and B are given by the roots of the equation x^2 - mx + 2m = 0

(iii) find the co-ordinates of m, the midpoint of AB

dont worry about (ii), i got it

now what about (iii)?

 

[Arithela]

another question:

Simplify (1/a-1) - (a/a-1)

is this simply = 1?

 

[SoulSearcher]

It is possible to do iii) without having to find the roots of the equation x² - mx + 2m = 0

Hopefully you remember that (x-a)(x-b) = x² - (a+b)x + ab, where a and b are the roots of the quadratic equation.

Using this knowledge, plus the information in part ii) of the question stating that the x-coordinates of A and B are the roots of the equation, you can thus determine that the x-coordinates of points A and B add up to m.

To show it in a more (hopefully clearer) way:
Let a and b be the x-coordinates of the points A and B respectively. [A: (a, y₁), B: (b, y₂)]
x² - mx + 2m = x² - (a+b)x + ab (the = is meant to be an equals sign with three horizontal lines, the congruency symbol, if that's the name for it)
Therefore a+b = m, equating corresponding terms in x

The midpoint of any two points on the Cartesian plane is given by [(x₁+x₂)/2, (y₁+y₂)/2]
Substituting the points A and B into this, it is clear that the x-coordinate of the midpoint of AB is given by (a+b)/2.
Since we know that a+b = m, we can thus determine that the x-coordinate of the midpoint of AB is m/2.

We know the x-coordinate of the midpoint, and now resorting to part i), we can now find the y-coordinate of the midpoint of AB, since the midpoint lies on the line L, which we have found the equation for in part i)

From part i),
y = mx - 2m; x = m/2, therefore
y = m(m/2) - 2m
= m² - 2m
= (m² - 4m)/2

Therefore the coordinates of the midpoint of AB is (m/2, (m² - 4m)/2)

 

[Arithela]

no it was a separate question...

this is what i did:

let roots be a and 1/a

a x 1/a = 1

a x 1/a = c/a

therefore c/a = 1

and then i solved and got an answer for the variable