Question

A quadratic equation is shown below:

25x2 + 10x + 1 = 0

Part A: Describe the solution(s) to the equation by just determining the radicand. Show your work. (5 points)

Part B: Solve 4x2 − 4x + 1 = 0 by using an appropriate method. Show the steps of your work, and explain why you chose the method used. (5 points)

Answer 1

@Hero

Answer 2

Part A just means to use \(b^2 - 4ac\) portion of the quadratic equation to determine if the quadratic is factorable.

Answer 3

can you explain what you mean

Answer 4

im confused

Answer 5

The general form of a quadratic expression is \(ax^2 + bx + c\) The \(a,b,c\) values from that can be plugged in to the quadratic formula
\(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Answer 6

ok

Answer 7

after you do dat den what e.e

Answer 8

Do you think you can identify the \(a\) \(b\) and \(c\) values from the first quadratic?

Answer 9

ummmmmm ill try

Answer 10

a=25x^2
b=10x
c=1

Answer 11

\(a, b, c\) are only the coefficients. just the numbers not the variables

Answer 12

a = 25, b = 10, c = 1

Answer 13

ok

Answer 14

Then what do we do once we have the numbers like that

Answer 15

We put them in the \(b^2 - 4ac\) portion of the quadratic equation and see if we get a positive or negative value.

Answer 16

so 10^2 - 4(25)(1)?

Answer 17

If \(b^2 - 4ac \ge 0\) then \(ax^2 + bx + c\) is factorable.
If \(b^2 - 4ac < 0\) then \(ax^2 + bx + c\) is not factorable.

Answer 18

^^ok so do i do it this way Look above ur answer

Answer 19

@Hero ^^

Answer 20

Yes, what was your result?

Answer 21

This always confused me sooo just a heads up xD

Answer 22

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Hero
Yes, what was your result?
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 23

idk i was waiting to see if it was yes or no so give me a sec plz

Answer 24

i got no clue i messed up some were or something cause im confused now

Answer 25

i got 0 as the answer

Answer 26

Very good.

Answer 27

So is Part A factorable?

Answer 28

No?

Answer 29

Why not?

Answer 30

Bc its not > greater than zero its < zero

Answer 31

What does it mean for \(b^2 - 4ac\) to be \(\ge\) to zero.

Answer 32

it can be factored

Answer 33

Yes, correct.

Answer 34

\(\color{#0cbb34}{\text{Originally Posted by}}\) hero
If \(b^2 - 4ac \ge 0\) then \(ax^2 + bx + c\) is factorable.
If \(b^2 - 4ac < 0\) then \(ax^2 + bx + c\) is not factorable.
\(\color{#0cbb34}{\text{End of Quote}}\)
like this but i thought since its zero it wouldn't be factorable

Answer 35

@Hero so it is factorable

Answer 36

@AnimeGhoul8863 Yes.

Answer 37

then what @Hero

Answer 38

@Hero You still there

Answer 39

@Shadow Can u pick up were hero left off

Answer 40

So @AnimeGhoul8863, if it is factorable, then the next step would be to make an attempt to factor the expression.

Answer 41

What do you mean by that

Answer 42

@Hero

Answer 43

im confused again

Answer 44

I've seen you factor quadratic expressions before. I'm surprised.

Answer 45

i think its bc i factor it when i see everything together but when its split up in multiple messages it confuses meh

Answer 46

@Hero you there you keep adding something and then leaves for like 10 mins

Answer 47

Sorry about that

Answer 48

We determined that \(25x^2 + 10x + 1\) was factorable and next step is to make an attempt to factor the expression. I've seen you factor in the past. Are you saying you don't have an approach to factoring this one?

Answer 49

25x^2+10x+1=
(25x^2 +5x) + (5x+1)
5x(5x+1)+(5x+1)
(5x+1)(5x+1)
(5x+1)^2

Answer 50

^This @Hero

Answer 51

Correct. The unique thing about when we get zero as the result of \(b^2 - 4ac\). When you factor \(ax^2 + bx + c\) you get a perfect square.

Answer 52

x'D @Hero Im sorry but can you help me word that we did the work but im confused how to put it all together

Answer 53

cause everything im trying to do to word this out isnt working

Answer 54

Can you put everything together for me so i can understand it and write it down please

Answer 55

@Hero

Answer 56

@Hero

Answer 57

All i need you to do is put everything together so i can write it down den were doneeeee

Answer 58

@Hero

Answer 59

e.e

Answer 60

To summarize:

The radicand is the number under the square root portion of the quadratic formula \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) which can be determined by using the expression \(b^2 - 4ac\).

If the value of \(b^2 - 4ac\) is \(> 0\), then \(ax^2 + bx + c\) is factorable.
If the value of \(b^2 - 4ac\) is \(= 0\), then \(ax^2 + bx + c\) is a perfect square.
If the value of \(b^2 - 4ac\) is \(< 0\), then \(ax^2 + bx + c\) is not factorable.

In this case, we take the \(a, b, \text{ and } c\) values from \(25x^2 + 10x + 1\) and plug them in to the radicand. So in this case \(a = 25, b = 10, c = 1\). Plugging that into the radicand we get
\(b^2 - 4ac = 10^2 - 4(25)(1) = 100 - 100 = 0\). So we know \(25x^2 + 10x + 1\) is a perfect square which we factor to get \((5x + 1)^2\).

But we're not done because we still have to find the solution to the quadratic equation since the original expression was set equal to zero. So in fact, we have to find the solution to \((5x + 1)^2 = 0\). Once we have found the values for x, then will be finished.

Answer 61

im confused Whats The answer for Part A lets start there

Answer 62

we have done so much this is getting jumbled up

Answer 63

Part A

Answer 64

take the a,b, and c values from 25x2+10x+1 and plug them in to the radicand.
a=25,b=10,c=1.
Plugging that into the radicand we get
b2−4ac=102−4(25)(1)=100−100=0.
25x2+10x+1 is a perfect square which we factor to get (5x+1)2.

Answer 65

What I said above is most of Part A.

Answer 66

All that was left was solving \((5x + 1)^2 = 0\) for \(x\).

Answer 67

ok stay here it will only take a sec

Answer 68

And then afterwards you discover something about the solution to an equation with perfect square binomial/trinomial.

Answer 69

x=-1over 5

Answer 70

Which is that there is only one solution. So the who point of part A was to get you to realize that the solution to \(25x^2 + 10x + 1 = 0\) has only one solution.

Answer 71

OMG its 10:00 PM AND WE HAVE TO DO PART B OMG T~T CAN WE TRY TO GO FAST WITH THIS PLZ

Answer 72

So, what you can say here for Part A is that since the value of the radicand for \(25x^2 + 10x + 1\), is \( 0\), then there will only be one solution for the equation \(25x^2 + 10x + 1 = 0\)

Answer 73

And then you say what that solution is.

Answer 74

Sorry that took so long to type. I don't know why

Answer 75

Part A: : take the a,b, and c values from 25x2+10x+1 and plug them in to the radicand.
a=25,b=10,c=1.
Plugging that into the radicand we get
b2−4ac=102−4(25)(1)=100−100=0.
25x2+10x+1 is a perfect square which we factor to get (5x+1)^2.
Once we have (5x+1)^2 we solve for x
x= -1 over 5
the solution to 25x2+10x+1=0 has only one solution

^^what i put

Answer 76

its ok lets just do part B

Answer 77

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Hero
So, what you can say here for Part A is that since the value of the radicand for \(25x^2 + 10x + 1\), is \( 0\), then there will only be one solution for the equation \(25x^2 + 10x + 1 = 0\)
\(\color{#0cbb34}{\text{End of Quote}}\)
Put this but include the solution to the equation with it. And show your solution steps.

Answer 78

You should show the steps you took to find the radicand and then show the steps you found to get the solution.

Answer 79

Actually, for Part B, you use everything you learned in Part A to do that on your own.

Answer 80

Part A: : 25x2+10x+1, is 0, 25x2+10x+1=0
take the a,b, and c values from 25x2+10x+1 and plug them in to the radicand.
a=25,b=10,c=1.
Plugging that into the radicand we get <step
b2−4ac=102−4(25)(1)=100−100=0.
25x2+10x+1 is a perfect square which we factor to get (5x+1)^2.<step
Once we have (5x+1)^2 we solve for x <step
x= -1 over 5
the solution to 25x2+10x+1=0 has only one solution


^^^the steps are there

Answer 81

OMG CAN YOU PLEASE JUST GIVE IT TOO ME IM DIEING HERE I CANT DO IT ON MY OWN THATS WHY IM HERE I JSUT WANNA GET THIS DONE SO PLEASE HELP ME GET THIS ALL DONE PLEASEEEEEEE

Answer 82

Yeah, but you have to include what you understand about getting zero as the value of the radicand.

Answer 83

its almost 10:30 PM can you just bend the rules alittle bit u showed me how to do it just put it all together so i can put it down and be done PLEASEEEE

Answer 84

@Shadow

Answer 85

@Hero Were did you gooooo

Answer 86

The logic used to solve Part A is the same logic you use to solve Part B.

Answer 87

and what is that logic called again

Answer 88

quadratic method?

Answer 89

Following the sequence of steps outlined here:

1. Determine the value of the radicand:

The radicand is the number under the square root portion of the quadratic formula \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) which can be determined by using the expression \(b^2 - 4ac\).

2. Determine from the value of the radicand whether the given quadratic expression is factorable:
If the value of \(b^2 - 4ac\) is \(> 0\), then \(ax^2 + bx + c\) is factorable.
If the value of \(b^2 - 4ac\) is \(= 0\), then \(ax^2 + bx + c\) is a perfect square.
If the value of \(b^2 - 4ac\) is \(< 0\), then \(ax^2 + bx + c\) is not factorable.

In this case, we take the \(a, b, \text{ and } c\) values from \(25x^2 + 10x + 1\) and plug them in to the radicand. So in this case \(a = 25, b = 10, c = 1\). Plugging that into the radicand we get
\(b^2 - 4ac = 10^2 - 4(25)(1) = 100 - 100 = 0\).

3. Factor the expression:
So we know \(25x^2 + 10x + 1\) is a perfect square which we factor to get \((5x + 1)^2\).

4. Find the solution to the original equation:
But we're not done because we still have to find the solution to the quadratic equation since the original expression was set equal to zero. So in fact, we have to find the solution to \((5x + 1)^2 = 0\). Once we have found the values for x, then will be finished.
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 90

Part A: the solution to 25x2+10x+1=0 has only one solution
take the a,b, and c values from 25x2+10x+1 and plug them in to the radicand.
a=25,b=10,c=1.
Plugging that into the radicand we get
b2−4ac=102−4(25)(1)=100−100=0.
25x2+10x+1 is a perfect square which we factor to get (5x+1)^2.
Once we have (5x+1)^2 we solve for x
x= -1 over 5
Part B: The method i used was quadratic method because it was easy to find the answer for part A with that method so i used the same method for part B.
4x2-4x+1=0
(22x2 - 4x) + 1 = 0
Factor 4x2-4x+1
4x2 - 2x - 2x - 1
2x * (2x-1)
(2x-1) * (2x-1)
Multiply (2x-1) by (2x-1)
(2x - 1)2 = 0
(2x-1)2 = 0 solve for x
x=1 over 2

Answer 91

^this is the best i can get it

Answer 92

Okay, as long as you can show your steps. Some of your steps look a bit puzzling though.

Answer 93

Part A: (52x2 + 10x) + 1 = 0
Factor 25x2+10x+1
25 * 1 = 25
5 + 5 = 10
25x2 + 5x + 5x + 1
5x * (5x+1)
(5x+1) * (5x+1)
(5x + 1)2 = 0
x = -1/5

Answer 94

^better

Answer 95

It would be better if you learned how to use \(LaTeX\)

Answer 96

Part A:the solution to 25x2+10x+1=0 has only one solution
(52x2 + 10x) + 1 = 0
Factor 25x2+10x+1
25 * 1 = 25
5 + 5 = 10
25x2 + 5x + 5x + 1
5x * (5x+1)
(5x+1) * (5x+1)
(5x + 1)2 = 0
x = -1/5

Part B: The method i used was quadratic method because it was easy to find the answer for part A with that method so i used the same method for part B.
4x2-4x+1=0
(22x2 - 4x) + 1 = 0
Factor 4x2-4x+1
4x2 - 2x - 2x - 1
2x * (2x-1)
(2x-1) * (2x-1)
Multiply (2x-1) by (2x-1)
(2x - 1)2 = 0
(2x-1)2 = 0 solve for x
x=1 over 2

Answer 97

FOR THE LAST TIME I CANT

Answer 98

You avoid using carets for exponents and other minor quirks

Answer 99

WHAT

Answer 100

It looks like @AnimeGhoul8863 is using factorization method

Answer 101

Part A:the solution to 25x2+10x+1=0 has only one solution
(52x2 + 10x) + 1 = 0
Factor 25x2+10x+1
25 * 1 = 25
5 + 5 = 10
25x2 + 5x + 5x + 1
5x * (5x+1)
(5x+1) * (5x+1)
(5x + 1)2 = 0
x = -1/5

Part B: The method i used was factorization method because it was easy to find the answer for part A with that method so i used the same method for part B.
4x2-4x+1=0
(22x2 - 4x) + 1 = 0
Factor 4x2-4x+1
4x2 - 2x - 2x - 1
2x * (2x-1)
(2x-1) * (2x-1)
Multiply (2x-1) by (2x-1)
(2x - 1)2 = 0
(2x-1)2 = 0 solve for x
x=1 over 2

Answer 102

@MARC I did pick up on that much. Just some of the steps are not clear but also, she neglects use of carets for exponents.

Answer 103

HERO IF U WANNA ADD THEM ID BE HAPPPY FOR U TO DO THAT BUT IF ITS NOT A HUGE DEAL I SHOULD BE FINE

Answer 104

ITS 10:30 AT NIGHT I JUST WANNA FINISH THIS

Answer 105

agree with u @Hero

Answer 106

@AnimeGhoul8863 I think you finished it to the best of your ability. The solutions are not wrong. All you need to work on is your ability to explain what you're doing at each step. Hopefully, what I posted above can help.

Answer 107

you told me i didnt have to OMG im just gonna submit it the way it is i cant add anymore

Answer 108

Thank you hero for your help but this is what im submitting

Answer 109

Part A:the solution to 25x2+10x+1=0 has only one solution
(52x2 + 10x) + 1 = 0
Factor 25x2+10x+1
25 * 1 = 25
5 + 5 = 10
25x2 + 5x + 5x + 1
5x * (5x+1)
(5x+1) * (5x+1)
(5x + 1)2 = 0
x = -1/5

Part B: The method i used was factorization method because it was easy to find the answer for part A with that method so i used the same method for part B.
4x2-4x+1=0
(22x2 - 4x) + 1 = 0
Factor 4x2-4x+1
4x2 - 2x - 2x - 1
2x * (2x-1)
(2x-1) * (2x-1)
Multiply (2x-1) by (2x-1)
(2x - 1)2 = 0
(2x-1)2 = 0 solve for x
x=1 over 2

Answer 110

^^what im submitting @Hero

Answer 111

Thank you for your help