What two numbers when added together give you -4/5?

- anonymous

What two numbers when added together give you -4/5?

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- asnaseer

just select any one number, say 10, and call the other one x, then write an equation to represent the conditions, i.e:
10 + x = -4/5
and solve to find x.

- anonymous

But that only gives me one number -54/5

- apoorvk

the other no. is 10 ofcourse!!!

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- asnaseer

the other one is the one we picked at random - i.e. 10

- apoorvk

the one you assumed in the first place!

- anonymous

Okay so my problem is -1/5x2-2x+4 and I factored this to -1/5x2+10x-54/5x+4. How do I factor out -1/5x2+10x and -54/5x+4?

- asnaseer

I don't understand what you are trying to do?

- anonymous

I need to factor -1/5x^2-2x+4 so I get an answer in ( - )^2 form.

- asnaseer

the equation you have given cannot be factored into something squared. it can, however, be factored into something squared minus something else.

- asnaseer

are you trying to solve a quadratic by completing the square?

- anonymous

Yes.

- asnaseer

ok, so the actual problem you are given is:\[-\frac{1}{5}x^2-2x+4=0\]if this is correct, then I would first simplify it by multiplying both sides by 5 to get rid of the fraction

- anonymous

f(x)=-1/5x2-2x+1/4 so I multiply both sides by 20?

- asnaseer

hang on - you have just changed the equation - is the last term "4" or "1/4"?

- anonymous

Sorry it is 1/4

- asnaseer

and what exactly does the question ask you to do?

- anonymous

Write in standard form of y(x)=a(x+h)^2+k... my notes say to get this you use the complete the square method. Is this correct?

- asnaseer

yes - but it is not what I originally thought was the question, which was to find the roots of the equation.
so here what you should do is notice first that the coefficient of x in a(x+h)^2+k is one. therefore you need to get your equation in that form. you can do this as follows:\[f(x)=-\frac{1}{5}x^2-2x+\frac{1}{4}=-\frac{1}{5}(x^2+10x-\frac{5}{4})\]

- asnaseer

then, since you need to get this into (x+h)^2, look at just the terms involving x^2 and x - here you have \(x^2+10x\), and see if you can write this in the form(x+h)^2 (ignoring any final constant).
try this and let me know what you get.

- asnaseer

i.e. try to find an h such that:\[(x+h)^2=x^2+10x+something\]

- anonymous

I'm sorry... how did you get the 10x?

- asnaseer

We are trying to get \(x^2+10x\) into a form \((x+h)^2\) because the equation for f(x) above involved these two as the first two terms, i.e.:\[f(x)=-\frac{1}{5}(x^2+10x-\frac{5}{4})\]

- anonymous

10x came from the -2x?

- asnaseer

yes

- asnaseer

\[-\frac{1}{5}\times10=-2\]

- asnaseer

I took \(-\frac{1}{5}\) out as common factor of the whole equation.

- anonymous

Okay so I get -1/5 (x^2+10x)=1/4.

- asnaseer

no

- anonymous

Sorry not =1/4 +1/4

- asnaseer

I think you are getting confused with this, let me try and start again...

- asnaseer

your original equation was given as:\[f(x)=-\frac{1}{5}x^2-2x+\frac{1}{4}\]
agreed?

- anonymous

Correct.

- asnaseer

ok, so next I took the -1/5 out as a common factor to get:\[f(x)=-\frac{1}{5}(x^2+10-\frac{5}{4})\]understand so far?

- asnaseer

sorry I missed the x in 10x

- asnaseer

\[f(x)=-\frac{1}{5}(x^2+10x-\frac{5}{4})\]

- asnaseer

with me so far?

- anonymous

Yes.

- asnaseer

ok, so next we see that within the brackets we have \(x^2+10x-\frac{5}{4}\).
and we know we need to get this in form \((x+h)^2\).
make sense?

- anonymous

This is where I get lost... what number is h?

- asnaseer

h is not known yet - that is what we need to find

- anonymous

And our x is 10 correct?

- asnaseer

no, x is an unknown of the equation

- anonymous

Okay so this is what x and h are what we are looking for next.

- asnaseer

all we are trying to do is rewrite \(x^2+10x-\frac{5}{4}\) in the form \((x+h)^2\). so we need o find a suitable value for h.
so what we do is first ignore the \(-\frac{5}{4}\)constant term, and just concentrate on \(x^2+10x\).
so we want to find a value for h such that:\[(x+h)^2=x^2+10x+something\]and we don't care what the \(something\) is.

- asnaseer

so, can you find a suitable value for h such that when we expand \((x+h)^2\) we get 10 as the coefficient of x?

- asnaseer

\[(x+h)^2=x^2+2hx+h^2\]so we need 2h=10, which means h must be 5.
understand?

- anonymous

Yes because were diving 2h/2=10/2 gives us 5.

- anonymous

dividing

- asnaseer

ok, so now we know h, we can get back to the original equation within the brackets, which was:\[x^2+10x-\frac{5}{4}\]and we know:\[(x+5)^2=x^2+10x+25\]therefore, we can write:\[x^2+10x-\frac{5}{4}=(x+5)^2-25-\frac{5}{4}\]

- asnaseer

the "-25" is there to get rid of the "+25" from the expansion of (x+5)^2

- asnaseer

agreed?

- anonymous

Yes.

- asnaseer

ok, so finally we plug this into the original equation to get:\[\begin{align}
f(x)&=-\frac{1}{5}(x^2+10x-\frac{5}{4})\\
&=-\frac{1}{5}((x+5)^2-25-\frac{5}{4})\\
&=-\frac{1}{5}(x+5)^2+5+\frac{1}{4}\\
&=-\frac{1}{5}(x+5)^2+5\frac{1}{4}\\
\end{align}\]

- asnaseer

and this is in the desired form of f(x)=a(x+h)^2+k

- asnaseer

I hope I explained it well enough for you to understand.

- anonymous

Wow. I was totally going to wrong way before. Thank you so much... This helps a lot. My teacher didnt explain the standard form in fractions so I was trying to go about it in another direction. I should be able to follow this example on the rest of my problems. Thanks again!

- asnaseer

yw - I'm glad I was able to help.

- anonymous

Will my axis of symmetry be 5, 5 1/4?

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