Melody

Let a and b be the solutions of the quadratic equation 2x^2 - 8x + 7 = 0. Find \frac{1}{2a} + \frac{1}{2b}.

\(\text{Let }\alpha\;\; and \;\ \beta\;\; \text{be the solutions of the quadratic equation }\\ 2x^2 - 8x + 7 = 0. \;\;\;Find \;\;\ \frac{1}{2a} + \frac{1}{2b}\\ \frac{1}{2\alpha} + \frac{1}{2\beta}=\frac{\beta + \alpha}{2\alpha\beta} \)

Now you can do this the long way and work out what the roots are but I expect you are supposed to know this:

\(\boxed{If \;\;ax^2+bx+c=0 \text{ and the roots are }\alpha \;\;and \;\; \beta\;\;then\\ \alpha + \beta = \frac{-b}{a}\;\;and \;\; \alpha \beta = \frac{c}{a}\\}~\\ \)

So for your question

\(\alpha + \beta = \frac{-b}{a}\;\;and \;\; \alpha \beta = \frac{c}{a}\\ \alpha + \beta = \frac{8}{2}\;\;and \;\; \alpha \beta = \frac{7}{2}\\\)

\(\frac{1}{2\alpha} + \frac{1}{2\beta}\\=\frac{\beta + \alpha}{2\alpha\beta}\\ =\frac{8}{2}\div\frac{2\times7}{2}\\ =4\div7\\ =\frac{4}{7}\)

Thanks Rosala and Manuel,

A very big thank you to both of you.  

I think I speak for all the answerers here :)

AND

Rosala, what beautiful flowers and chocolate cake and I get a bit of both. 

Thanks  

Perhaps someone else would like to help with this explanation...??

Any takers?

Here is a different example:

let

\(f(x)=x^2\)

This is a concave up parabola with vertex (0,0)

Now \(f(2x) = (2x)^2\\f(2x) = 4x^2\)

How does the second one differ from the first.

Here is the graphs

The green one is f(x) and the blue one is f(2x)

Can you see that for any given y value the new corresponding x is half the size.

So when y=36 the x value was originally \(\pm6\)     \(If \;f(x)=x^2 \;\;then \;\;f(6)=36\)

But when x is replaced with 2x then  it becomes   \( f(2x) =4x^2\)  as show above and  \(f(3)=36\)

so if you double the x in the equation then the resulting graph will have half the width!

Hi all,

Supermanaccz has asked me to come back an talk about this ancient question.

Thanks for your interest Supermanaccz    

Firstly the question is not displaying properly because the coding used on this site has changed a lot in the past 3 and a bit years.

So I will try to put the question up how it may originally have been. I believe a diagram like the one below would have been included in the original question.  Thie is the graph of   y=f(x).

I have drawn it in a program called GeoGebra. GeoGebra is a great tool. I found it a little hard to use at first (especially for more complicated things) but if you persist with it it will become much easier. Geogebra is a free download. I use it often.

Now to the actual question:

The graph of  is shown below. Assume the domain of  is  [-4,4] and that the vertical spacing of grid lines is the same as the horizontal spacing of grid lines. 

Part (a): The points  (a, 4) and  (b, -4)  are on the graph of  y=f(2x)    Find  a  and  b

From the graph you can see that 2 points on the f(x)  graph are (-4,4) and (4,-4)

-------------------------------------

Here is what i said before:

Part a       y=f(2a)

f(-4)=4   but    f(2a)=-4    so   2a=-4     so   a=-2

f(4)=-4   but    f(2b)=4     so    2b=4      so    b=2

----------------------------------------

But in the next post I will do an entirley different example, perhaps you will be able to see how many graphs are related to parent graphs. In this case f(x) is the parent function and f(2x) it the 'offspring'  Multiplying the x by 2 will cause the graph to have half the width (from the y axis). 

Part (b): Find the graph of  y=2x         Verify that your points from part (a) are on the graph. 

Part (c): The points (c,4)  and (d,-4)  are on the graph of y=f(2x-8)     Find  c   and  d

Part (d): Find the graph of  y=f(2x-8)   Be sure to verify that your points from part (c) are on the graph both algebraically and geometrically. 

2C1*0.6*0.4=2*0.6*0.4=0.48

This was an easy to type qeustion

SO

Next time could you please just type it in!

Mr. Smith brings home 7 animals for his 7 children. Each child will adopt a pet to be her or his own. There are 4 different cats (a Siamese, a Persian, a Calico, and a Minx), 2 different dogs (a Poodle and a Golden Retriever), and a goldfish. Anna and Betty refuse to take care of the goldfish, and Charlie and Danny insist on having cats. The other 3 kids are easier to please -- they'll take anything. In how many ways can Mr. Smith give the children pets they'll want?

First we must make sense of this:

cat1, cat2, cat3, cat4,      Charlie and Danny and 2 others

Dog1, dog2,   

GoldFish                          NOT  Anna or Betty

I am going to attmept to allocat the pets in this order.

Charlie (4 cats), Danny (3 cat possibliities), Anna (4 non-fish pets left), Betty (3 non-fish left) , C5 (3 pets left), C6(2 pets left), C7 (this child gets what is left)

4*3*4*3*3*2*1 = 864 ways to allocate the pets.

I count 5

4 at right angles to the square end and one more.

Well for starters you stop giving other people thumbs down for no reason !!