anonymous
  • anonymous
What two numbers when added together give you -4/5?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
asnaseer
  • 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
  • anonymous
But that only gives me one number -54/5
apoorvk
  • apoorvk
the other no. is 10 ofcourse!!!

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

asnaseer
  • asnaseer
the other one is the one we picked at random - i.e. 10
apoorvk
  • apoorvk
the one you assumed in the first place!
anonymous
  • 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
  • asnaseer
I don't understand what you are trying to do?
anonymous
  • anonymous
I need to factor -1/5x^2-2x+4 so I get an answer in ( - )^2 form.
asnaseer
  • asnaseer
the equation you have given cannot be factored into something squared. it can, however, be factored into something squared minus something else.
asnaseer
  • asnaseer
are you trying to solve a quadratic by completing the square?
anonymous
  • anonymous
Yes.
asnaseer
  • 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
  • anonymous
f(x)=-1/5x2-2x+1/4 so I multiply both sides by 20?
asnaseer
  • asnaseer
hang on - you have just changed the equation - is the last term "4" or "1/4"?
anonymous
  • anonymous
Sorry it is 1/4
asnaseer
  • asnaseer
and what exactly does the question ask you to do?
anonymous
  • 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
  • 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
  • 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
  • asnaseer
i.e. try to find an h such that:\[(x+h)^2=x^2+10x+something\]
anonymous
  • anonymous
I'm sorry... how did you get the 10x?
asnaseer
  • 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
  • anonymous
10x came from the -2x?
asnaseer
  • asnaseer
yes
asnaseer
  • asnaseer
\[-\frac{1}{5}\times10=-2\]
asnaseer
  • asnaseer
I took \(-\frac{1}{5}\) out as common factor of the whole equation.
anonymous
  • anonymous
Okay so I get -1/5 (x^2+10x)=1/4.
asnaseer
  • asnaseer
no
anonymous
  • anonymous
Sorry not =1/4 +1/4
asnaseer
  • asnaseer
I think you are getting confused with this, let me try and start again...
asnaseer
  • asnaseer
your original equation was given as:\[f(x)=-\frac{1}{5}x^2-2x+\frac{1}{4}\] agreed?
anonymous
  • anonymous
Correct.
asnaseer
  • 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
  • asnaseer
sorry I missed the x in 10x
asnaseer
  • asnaseer
\[f(x)=-\frac{1}{5}(x^2+10x-\frac{5}{4})\]
asnaseer
  • asnaseer
with me so far?
anonymous
  • anonymous
Yes.
asnaseer
  • 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
  • anonymous
This is where I get lost... what number is h?
asnaseer
  • asnaseer
h is not known yet - that is what we need to find
anonymous
  • anonymous
And our x is 10 correct?
asnaseer
  • asnaseer
no, x is an unknown of the equation
anonymous
  • anonymous
Okay so this is what x and h are what we are looking for next.
asnaseer
  • 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
  • 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
  • asnaseer
\[(x+h)^2=x^2+2hx+h^2\]so we need 2h=10, which means h must be 5. understand?
anonymous
  • anonymous
Yes because were diving 2h/2=10/2 gives us 5.
anonymous
  • anonymous
dividing
asnaseer
  • 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
  • asnaseer
the "-25" is there to get rid of the "+25" from the expansion of (x+5)^2
asnaseer
  • asnaseer
agreed?
anonymous
  • anonymous
Yes.
asnaseer
  • 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
  • asnaseer
and this is in the desired form of f(x)=a(x+h)^2+k
asnaseer
  • asnaseer
I hope I explained it well enough for you to understand.
anonymous
  • 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
  • asnaseer
yw - I'm glad I was able to help.
anonymous
  • anonymous
Will my axis of symmetry be 5, 5 1/4?

Looking for something else?

Not the answer you are looking for? Search for more explanations.