Question

How to solve \[2x^2 -10x + 3 = 0\] by completing the square?

Answer 1

what do you think so far?

Answer 2

Well, I was thinking to first divide both sides by 2, so you can have x^2 instead of 2x^2

Answer 3

use the quadratic formula instead, it is easier in this case

Answer 4

^^

Answer 5

yeah its easier

Answer 6

I would, but the question asks to complete the square cx

Answer 7

it makes no difference
the formula is identical to completing the square

Answer 8

just does it as a formula instead of in steps

Answer 9

do it as a foruma which is what @satellite73 said im gonna go help some other people looks like you got all the help you need @camerondoherty

Answer 10

\[2x^2-10x=3\] is a start, then divide by 2
\[x^2-5x=\frac{3}{2}\] and now with annoying fractions go right to
\[(x-\frac{5}{2})^2=\frac{3}{2}+\frac{9}{4}\]

Answer 11

\(\large\color{blue}{ 2x^2-10+3=0 }\)

\(\large\color{blue}{ 2x^2-10+3\color{red}{-3}=0\color{red}{-3} }\)

\(\large\color{blue}{ 2x^2-10=-3 }\)

\(\large\color{blue}{ 2(x^2-5)=-3 }\)

\(\large\color{blue}{ 2(x^2-5\color{green}{+\frac{\LARGE 5^2}{\LARGE 2^2}}\color{red}{-\frac{\LARGE 5^2}{\LARGE 2^2}})=-3 }\)

\(\large\color{blue}{ 2(x^2-5\color{green}{+\frac{\LARGE 5^2}{\LARGE 2^2}})-2\color{red}{\frac{\LARGE 5^2}{\LARGE 2^2}}=-3 }\)

Answer 12

2x^2 -10x + 3 = 0
2x2−10x+3=0
Step 1: Use quadratic formula with a=2, b=-10, c=3.
x=
−b±sqrt(b2−4ac)
2a
x=
−(−10)±sqrt((−10)2−4(2)(3))
2(2)
x=
10±sqrt(76)
4
x=4.6794494717703365,0.320550528229663
Answer:
x=4.6794494717703365,0.320550528229663

Answer 13

2 many, lol

Answer 14

oh whoops it is
\[2x^2-10x+3=0\] so really \[(x-\frac{5}{2})^2=-\frac{3}{2}+\frac{9}{4}\]

Answer 15

solution is x=1/2(5-+sqr19

Answer 16

Hm, so
\[(x-\frac{5}{2})^2=-\frac{3}{2}+\frac{9}{4}\]
Adding the fractions on the side
\[(x-\frac{5}{2})^2=\frac{3}{4}\]
Then solving the other side
\[x^2-5x+\frac{ 25 }{ 4 }= \frac{ 3 }{ 4 }\]
I'm stuck here.

Answer 17

perfect square trinomial

Answer 18

the left side

Answer 19

\[x^2-5x+\frac{ 22 }{ 4 }= 0\]

Answer 20

what you had before is correct.

\(\large\color{blue}{(a+b)^2=a^2+2ab+b^2 }\)

\(\large\color{blue}{(a+b)^2=x^2+2(x)(5/2)+(5/2)^2 }\)

Answer 21

Oh left side. Got left/right confused, lol

Answer 22

See, compare it to that

Answer 23

\(\large\color{blue}{x^2+5x+25/4=3/4 }\)

is really,

\(\large\color{blue}{x^2+5(x)(5/2)+(5/2)^2=3/4 }\)

Answer 24

\(\large\color{blue}{a^2+2(a)(b)+(b)^2=>~(a+b)^2 }\)

Answer 25

\[ax^2+bx+c=0 \\ \text{ first step for your problem: subtract } c \text{ on both sides } \\ ax^2+bx=-c \\ \text{ second step for your problem: factor } a \text{ from both terms on left side } \\ \text{ but don't freak out because there isn't an } a \text{ next to the } bx \\ \text{ because you can put an } a \text{ there as long as you also divide by } a \text{ next to it also } \\ ax^2+\frac{a}{a}bx=-c \\ \text{ so factor out } a \text{ from both terms on the left } \\ a(x^2+\frac{1}{a}bx)=-c \\ a(x^2+\frac{b}{a}x)=-c\] \[\text{ third step: inside the parenthesis on the left side } \text{ add in a number that will complete the square } \\ \text{ but whatever you do to one side do to the other } \]
\[a(x^2+\frac{b}{a}x+(\frac{b}{2a})^2)=-c+a(\frac{b}{2a})^2\]
\[\text{ notice inside the ( ) I just added in the square of the number next to } x \text{ divided by 2 }\]
\[\text{ but that number I added in was being multiplied by } a \]
\[\text{ So I actually added on the left } a(\frac{b}{2a})^2\]
\[\text{ So I also need to add that number same number on the right }\]
So that is what I have above:
\[a(x^2+\frac{b}{a}x+(\frac{b}{2a})^2)=-c+a(\frac{b}{2a})^2\]
\[\text{ Now the whole reason I did this was so I could } \\ \text{ write that number on the left next to the number } a \text{ as something squared }\]
\[\text{ so we have: } a(x+\frac{b}{2a})^2=-c+a(\frac{b}{2a})^2\]
\[\text{ now I'm going to divide both sides by } a \]
\[(x+\frac{b}{2a})^2=\frac{-c}{a}+\frac{a}{a}(\frac{b}{2a})^2\]
\[(x+\frac{b}{2a})^2=\frac{-c}{a}+(\frac{b}{2a})^2\]
I'm going to go ahead and write the right hand side as one fraction:
\[(x+\frac{b}{2a})^2=\frac{-c}{a}+\frac{b^2}{4a^2}\]
\[(x+\frac{b}{2a})^2=\frac{-c(4a)}{a(4a)}+\frac{b^2}{4a^2}\]
\[(x+\frac{b}{2a})^2=\frac{-4ac+b^2}{4a^2}\]
\[(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}\]
Now recall the objective is to isolate x.
But oh no it is inside some square thingy...
So let's take the square root of both sides :
\[(x+\frac{b}{2a})=\pm \sqrt{ \frac{b^2-4ac}{4a^2}}\]
\[x+\frac{b}{2a}=\pm \sqrt{ \frac{b^2-4ac}{4a^2}}\]
\[x+\frac{b}{2a}= \pm \frac{\sqrt{b^2-4ac}}{\sqrt{4a^2}}\]
\[x+\frac{b}{2a}= \pm \frac{\sqrt{b^2-4ac}}{\pm 2 a}\]
We don't need plus or minus twice:
\[x+\frac{b}{2a}= \pm \frac{\sqrt{b^2-4ac}}{ 2a} \]
Subtracting b/2a on both sides give:
\[x=\frac{-b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}\]
And you can combine the fractions here and thus the quadratic formula is born.
\[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]
Now the reason I wrote this...
Is because if you can understand how to complete the square with just "letters" involved then you can do it for any numbers these "letters" can represent.
\[\text{ Also I hope you notice } a \text{ cannot be zero }\]
:)

Answer 26

the old equation editor, but very nice!

Answer 27

`\(\large\color{slate}{` .... `}\)`

Answer 28

anyways, back to math

Answer 29

wow you just proved the quadratic equation @freckles

Answer 30

we need completing the sq/ tho

Answer 31

well, not a lover of abstract math...

Answer 32

Well the method I show to get the quadratic formula involves completing the square...
If you can understand all of those steps I took with "letters" then he can do it with the numbers those "letters" can represent.

Answer 33

Sorry, The equation button went below and I couldnt reach it
I have to rewrite all of this cx
In my opinion, that may as well be a tutorial, its great enough to be c:
Thank you for writing it *u*
It really helps me understand!

So, Im not going to write in in latex, but I'm going to solve it on paper c:
Last time I tried writing it in latex, I lost it all, so give me a sec...

Answer 34

Retyped because some of the lines got cut off.
\[ax^2+bx+c=0 \\ \text{ first step for your problem: subtract } c \text{ on both sides } \\ ax^2+bx=-c \\ \text{ second step for your problem: factor } a \text{ from both terms on left side } \\ \text{ but don't freak out because there isn't an } a \text{ next to the } bx \\ \text{ because you can put an } a \text{ there as long as you also divide by } a \\ \text{ next to it also } \\ ax^2+\frac{a}{a}bx=-c \\ \text{ so factor out } a \text{ from both terms on the left } \\ a(x^2+\frac{1}{a}bx)=-c \\ a(x^2+\frac{b}{a}x)=-c \\
\text{ third step: inside the parenthesis on the left side } \\ \text{ add in a number that will } \\ \text{ complete the square but whatever you do to one side do to the other }


\\ a(x^2+\frac{b}{a}x+(\frac{b}{2a})^2)=-c+a(\frac{b}{2a})^2

\\ \text{ notice inside the ( ) I just added in the square of the number next to } \\ x \text{ divided by 2 }
\\ \text{ but that number I added in was being multiplied by } a \\ \text{ So I actually added on the left } a(\frac{b}{2a})^2 \\ \text{ so I need to add } a(\frac{b}{2a})^2 \text{ on the right as well } \\ \text{ So that is what I have above: } \\ a(x^2+\frac{b}{a}x+(\frac{b}{2a})^2)=-c+a(\frac{b}{2a})^2 \\ \text{ Now the whole reason I did this was so I could } \\ \text{ write that number on the left next to the number } a \text{ as something squared } \\ \text{ so we have: } a(x+\frac{b}{2a})^2=-c+a(\frac{b}{2a})^2 \\ \text{ now I'm going to divide both sides by } a \\ (x+\frac{b}{2a})^2=\frac{-c}{a}+\frac{a}{a}(\frac{b}{2a})^2 \\ (x+\frac{b}{2a})^2=\frac{-c}{a}+(\frac{b}{2a})^2 \\ \text{ I'm going to go ahead and write the right hand side as one fraction: } \\ (x+\frac{b}{2a})^2=\frac{-c}{a}+\frac{b^2}{4a^2} \\ (x+\frac{b}{2a})^2=\frac{-c(4a)}{a(4a)}+\frac{b^2}{4a^2} \\ (x+\frac{b}{2a})^2=\frac{-4ac+b^2}{4a^2} \\ (x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2} \\ \text{ Now recall the objective is to isolate x.
} \\ \text{ But oh no it is inside some square thingy...
} \\ \text{ So let's take the square root of both sides :
} \\ (x+\frac{b}{2a})=\pm \sqrt{ \frac{b^2-4ac}{4a^2}} \\
x+\frac{b}{2a}=\pm \sqrt{ \frac{b^2-4ac}{4a^2}} \\
x+\frac{b}{2a}= \pm \frac{\sqrt{b^2-4ac}}{\sqrt{4a^2}} \\ x+\frac{b}{2a}= \pm \frac{\sqrt{b^2-4ac}}{ 2a} \\
\text{ Subtracting } \frac{ b}{2a } \text{ on both sides give: } \\ x=\frac{-b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a} \\
\text{ And you can combine the fractions here and thus } \\ \text{ the quadratic formula is born. }\\ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}
\\ \text{ Now the reason I wrote this...
} \\ \text{ is because if you can understand how to } \\ \text{ complete the square with just "letters" involved } \\ \text{ then you can do it for any numbers these "letters" can represent. } \\ \text{ Also I hope you notice } a \text{ cannot be zero } \]

Answer 35

You re-wrote it? I kinda feel bad now cx
Thank You so much...
Omg
I don't know how I can thank you enough
You showed it just so thouroghly and so easy to understand!! c:

Answer 36

When I solved by plugging in the values, I got
\[\begin{array}{*{20}c} {x = \frac{{ 5 \pm \sqrt {3} }}{{2}}} & \\ \end{array}\]

Answer 37

\[2x^2 -10x + 3 = 0 \\
2x^2-10x=-3 \\
2(x^2-5x)=-3 \\
2(x^2-5x+(\frac{5}{2})^2)=-3+2(\frac{5}{2})^2 \\
\text{ now remember we actually added in }
2 \cdot (\frac{5}{2})^2
\\ \text{ on the left } \\
\text{ so we also needed to have added }
2 \cdot (\frac{5}{2})^2
\\
\text{ so on the left next to the 2 we can write that as something squared now }
\\ 2(x-\frac{5}{2})^2=-3+2 \cdot (\frac{5}{2})^2\]

Answer 38

I think we are suppose to get something different in the square root

Answer 39

like i think instead of 3 you should have something else there

Answer 40

So, I did something wrong then cx
Let me do the math again, one sec cx

Answer 41

ok

Answer 42

you can always use quadratic formula to check
\[x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2 }\]
\[x \frac{ 10\pm \sqrt{100-4(2)(3)} }{ 4 }\]

Answer 43

I multiplied wrong, cx
It's supposed to be:
\[\begin{array}{*{20}c} {x = \frac{{ 5 \pm \sqrt 19} }}{{2a}}} & \\ \end{array}\]

Answer 44

Wait, er, Broken Latex
\[\begin{array}{*{20}c} {x = \frac{{ - b \pm \sqrt {19} }}{{2a}}} \\ \end{array}\]

Answer 45

Wait, no cx
\[\begin{array}{*{20}c} {x = \frac{{ 5 \pm \sqrt {19} }}{{2}}} \\ \end{array}\]

Answer 46

that looks right to me

Answer 47

\[2x^2 -10x + 3 = 0 \\
2x^2-10x=-3 \\
2(x^2-5x)=-3 \\
2(x^2-5x+(\frac{5}{2})^2)=-3+2(\frac{5}{2})^2 \\
\text{ now remember we actually added in }
2 \cdot (\frac{5}{2})^2
\\ \text{ on the left } \\
\text{ so we also needed to have added }
2 \cdot (\frac{5}{2})^2
\\
\text{ so on the left next to the 2 we can write that as something squared now }
\\ 2(x-\frac{5}{2})^2=-3+2 \cdot (\frac{5}{2})^2 \\ 2(x-\frac{5}{2})^2=-3+2 \cdot \frac{25}{4} \\ \text{ divide both sides by 2 } \\ \frac{2}{2}(x-\frac{5}{2})^2=\frac{-3}{2}+ \frac{2}{2} \cdot \frac{25}{4} \\ (x-\frac{5}{2})^2=\frac{-3}{2}+\frac{25}{4} \]
then combine fractions on left
then take square root of both sides

Answer 48

combine fractions on right*

Answer 49

yes

Answer 50

from the quadratic formula