anonymous
  • anonymous
1) When the sum of 5 and twice a positive number is subtracted from the square of the number, 0 results. Find the number.
Mathematics
  • Stacey Warren - Expert brainly.com
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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.
schrodinger
  • schrodinger
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pooja195
  • pooja195
Let x be the number First thing it says is subtracted from the square of A NUMER (x) sooo \[\huge~\rm~x^2- \]
pooja195
  • pooja195
*number
pooja195
  • pooja195
Then it says the number is twice postivie soo 2x and the sum (adding) of 5 so it would look like this \[\huge~\rm~(2x+5)\]

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More answers

pooja195
  • pooja195
The result is 0 so =0 \[\huge~\rm~x^2-(2x+5)=0 \]
pooja195
  • pooja195
We can remove the parentheses and write it like this: \[\huge~\rm~x^2-2x-5=0 \]
pooja195
  • pooja195
Find the abc values :) a ,b , c values are \[\huge\rm Ax^2+Bx+C=0\] where a =leading coefficient b= leading coefficient c= constant term
anonymous
  • anonymous
a=1 b=2 c=5
anonymous
  • anonymous
-2 -5
pooja195
  • pooja195
Good : )
pooja195
  • pooja195
\[\huge~\rm~x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]
anonymous
  • anonymous
\[x=\frac{ 2\pm \sqrt{2^2-4(1)(5)} }{ 2(1) }\]
pooja195
  • pooja195
\[\huge~\rm~x=\frac{ -(-2) \pm \sqrt{(-2)^2-4(1)(-5)} }{ 2(1) }\]
anonymous
  • anonymous
oh in the sqrt its -2
anonymous
  • anonymous
and -5 okay
pooja195
  • pooja195
ok can you solve whats in the sqrt?
anonymous
  • anonymous
\[\sqrt{80}\]
anonymous
  • anonymous
\[2\pm4\sqrt{5}\]
pooja195
  • pooja195
I got \[\huge~\rm~1±1\sqrt{6}\]
anonymous
  • anonymous
..how?
pooja195
  • pooja195
eh i cheated and used a calc ;p heres the work http://prntscr.com/80bm1y
anonymous
  • anonymous
what calc did you use
pooja195
  • pooja195
http://www.mathwarehouse.com/quadratic/quadratic-formula-calculator.php
pooja195
  • pooja195
Then we make 2 seperate equations one plus and one minus since we basically have to split the plus minus sign then u simplfy from here \[\frac{ \huge~\rm~1+1\sqrt{6} }{ \huge~\rm~1 }\] \[\frac{ \huge~\rm~1-1\sqrt{6} }{ \huge~\rm~1 }\]
pooja195
  • pooja195
You will end up with ONE positive answer and ONE negative answer. The psotive answer is the answer to the question :)
anonymous
  • anonymous
...
pooja195
  • pooja195
...?
anonymous
  • anonymous
how
pooja195
  • pooja195
I never learned....:/ you will need to use a calculator or ask someone else...i have always used a calc for this part... @Astrophysics
ali2x2
  • ali2x2
@ganeshie8
just_one_last_goodbye
  • just_one_last_goodbye
@pooja195 what do we want to convert into a decimal?
pooja195
  • pooja195
\(\color{#0cbb34}{\text{Originally Posted by}}\) @pooja195 Then we make 2 seperate equations one plus and one minus since we basically have to split the plus minus sign then u simplfy from here \[\frac{ \huge~\rm~1+1\sqrt{6} }{ \huge~\rm~1 }\] \[\frac{ \huge~\rm~1-1\sqrt{6} }{ \huge~\rm~1 }\] \(\color{#0cbb34}{\text{End of Quote}}\) This and show your stpes
pooja195
  • pooja195
*steps
just_one_last_goodbye
  • just_one_last_goodbye
alright
just_one_last_goodbye
  • just_one_last_goodbye
both on top of each other like fractions correct?
ali2x2
  • ali2x2
@just_one_last_goodbye i think so
pooja195
  • pooja195
Actually a better way to solve it Apply this rule \[\huge~\rm~\frac{ a }{ 1 }=a\]
pooja195
  • pooja195
\[\huge~\rm~\frac{1+\sqrt{6}1}{1}=1+\sqrt{6}1 \]
pooja195
  • pooja195
\[\huge~\rm~1+1\cdot \sqrt{6} \]
pooja195
  • pooja195
\[\huge~\rm~1\cdot \:a=a\]
pooja195
  • pooja195
\[\huge~\rm~1+\sqrt{6}\]
pooja195
  • pooja195
Now turn it into decimal form
pooja195
  • pooja195
We dont even need to solve fore the subtraction problem because we know that the number will be negative
anonymous
  • anonymous
ok
ali2x2
  • ali2x2
ill try it myself
just_one_last_goodbye
  • just_one_last_goodbye
I got 3.44949
pooja195
  • pooja195
Thats right
anonymous
  • anonymous
I was gonna ask if we round it to 3.45
pooja195
  • pooja195
Yep :)
ali2x2
  • ali2x2
i got 3.44949
anonymous
  • anonymous
okay, thank you for being so patient.
ali2x2
  • ali2x2
oh they already said that
pooja195
  • pooja195
lol
ali2x2
  • ali2x2
i cri

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