Question

1) When the sum of 5 and twice a positive number is subtracted from the square of the number, 0 results. Find the number.

Answer 1

Let x be the number
First thing it says is subtracted from the square of A NUMER (x) sooo
\[\huge~\rm~x^2- \]

Answer 2

*number

Answer 3

Then it says the number is twice postivie soo 2x and the sum (adding) of 5 so it would look like this
\[\huge~\rm~(2x+5)\]

Answer 4

The result is 0 so =0
\[\huge~\rm~x^2-(2x+5)=0 \]

Answer 5

We can remove the parentheses and write it like this:
\[\huge~\rm~x^2-2x-5=0 \]

Answer 6

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

Answer 7

a=1
b=2
c=5

Answer 8

-2
-5

Answer 9

Good : )

Answer 10

\[\huge~\rm~x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]

Answer 11

\[x=\frac{ 2\pm \sqrt{2^2-4(1)(5)} }{ 2(1) }\]

Answer 12

\[\huge~\rm~x=\frac{ -(-2) \pm \sqrt{(-2)^2-4(1)(-5)} }{ 2(1) }\]

Answer 13

oh in the sqrt its -2

Answer 14

and -5 okay

Answer 15

ok can you solve whats in the sqrt?

Answer 16

\[\sqrt{80}\]

Answer 17

\[2\pm4\sqrt{5}\]

Answer 18

I got \[\huge~\rm~1±1\sqrt{6}\]

Answer 19

..how?

Answer 20

eh i cheated and used a calc ;p heres the work
http://prntscr.com/80bm1y

Answer 21

what calc did you use

Answer 22

http://www.mathwarehouse.com/quadratic/quadratic-formula-calculator.php

Answer 23

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
\[\frac{ \huge~\rm~1+1\sqrt{6} }{ \huge~\rm~1 }\]
\[\frac{ \huge~\rm~1-1\sqrt{6} }{ \huge~\rm~1 }\]

Answer 24

You will end up with ONE positive answer and ONE negative answer. The psotive answer is the answer to the question :)

Answer 25

...

Answer 26

...?

Answer 27

how

Answer 28

I never learned....:/ you will need to use a calculator or ask someone else...i have always used a calc for this part...
@Astrophysics

Answer 29

@ganeshie8

Answer 30

@pooja195 what do we want to convert into a decimal?

Answer 31

\(\color{#0cbb34}{\text{Originally Posted by}}\) @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

Answer 32

*steps

Answer 33

alright

Answer 34

both on top of each other like fractions correct?

Answer 35

@just_one_last_goodbye i think so

Answer 36

Actually a better way to solve it
Apply this rule
\[\huge~\rm~\frac{ a }{ 1 }=a\]

Answer 37

\[\huge~\rm~\frac{1+\sqrt{6}1}{1}=1+\sqrt{6}1 \]

Answer 38

\[\huge~\rm~1+1\cdot \sqrt{6} \]

Answer 39

\[\huge~\rm~1\cdot \:a=a\]

Answer 40

\[\huge~\rm~1+\sqrt{6}\]

Answer 41

Now turn it into decimal form

Answer 42

We dont even need to solve fore the subtraction problem because we know that the number will be negative

Answer 43

ok

Answer 44

ill try it myself

Answer 45

I got 3.44949

Answer 46

Thats right

Answer 47

I was gonna ask if we round it to 3.45

Answer 48

Yep :)

Answer 49

i got 3.44949

Answer 50

okay, thank you for being so patient.

Answer 51

oh they already said that

Answer 52

lol

Answer 53

i cri