Question

how to complete the square for this problem

Answer 1

\[3x^{2}+5x+1 =0\]

Answer 2

I know that I gotta get it in the form ax^2+bx = -c

Answer 3

\[3x^{2}+5x = -1 \]

Answer 4

then multiply by 1/a

\[\frac{ 1 }{ 3 }(3x^{2}+5x = -1)\]

\[x^{2}+\frac{ 5 }{ 3 }x = -\frac{ 1 }{ 3 }\]

just wondering what do I do after this?

Answer 5

Good work so far.
You are correct that the x^2 term must have a coefficient of 1 in order to complete the square. You have already done that.
Now take the coefficient of the x-term. Divide it by 2, and square it.
That is what you add to both sides.

Answer 6

thank you so I gotta do this right?

\[(\frac{ 5 }{ 3 }\div2)^{2}\]

Answer 7

Correct.

Answer 8

ok cool

Answer 9

Remember that to halve a fraction, just double the denominator.
Half of \(\dfrac{1}{a} \) is \(\dfrac{1}{2a} \).

Answer 10

\[3x^{2}+5x+1 =0~becomes~3(x^2+\frac{ 5 }{ 3 }x + \frac{ 1 }{ 3 }).\]

Answer 11

that's a neat trick

Answer 12

Foc\[x^2+\frac{ 5 }{ 3 }x.\]us on completing the square of

Answer 13

Half of \(\dfrac{1}{a} = \dfrac{1}{2} \times \dfrac{1}{a} = \dfrac{1}{2a} \)

Answer 14

a) Take half of the coefficient of x
b) Square that result
c) Add this latest result to \[x^2+\frac{ 5 }{ 3 }x\] and then subtract the same thing.
d) rewrite the first 3 terms as a perfect square.
e) simplify your final result as much as possible.

Answer 15

@mathmale before I was thinking that it had to be like this

\[\frac{ 5 }{ 3 } \div 1/2 = (\frac{ 5 }{ 3 }*2 )^{2}\]

Answer 16

No.

Answer 17

You need to divide 5/3 by 2.
That is simply 5/6.

Answer 18

\(\dfrac{5}{3} \div 2 = \dfrac{5}{3} \times \dfrac{1}{2} = \dfrac{5}{6} \)

Answer 19

No. \[\frac{ 5 }{ 3 } \div 1/2 ~should~be~\frac{ 5 }{ 3 }*\frac{ 1 }{ 2 },\]

Answer 20

Remember the rule of dividing fractions by multiplying by the reciprocal.

Answer 21

...which comes out to 5/6, as mathstudent55 has aptly said.

Answer 22

oh yeah I see that now thanks I got confused before

Answer 23

Take the coefficient of the x-term: 5/3
Divide it by 2: 5/6
Now square 5/6.
What do you get?

Answer 24

\[(x+\frac{ 5 }{ 6 })^{2} = \frac{ 13 }{ 36 }\]

Answer 25

Now square this (5/6). Add this square to your previous result, and then subtract it.
Please show all your work so that mathstudent55 and I can give you pertinent feedback.

Answer 26

this is what I got guys

Answer 27

\[(x+\frac{ 5 }{ 6 })^{2} = \frac{ 13 }{ 36 }\]

Answer 28

nice!

Answer 29

remember that we factored out that coefficient of 3. Aren't you going to include that 3 in your final result?

You can check your own work by expanding that square of the binomial (x+5/6). If doing this, and simplifying, takes you back to the very first expression you posted, you're right. If not, then it's back to the drawing board.

Answer 30

Square 5/6 to get 25/36.
Now add 25/36 to both sides.

\(x^2 + \dfrac{5}{3} x + \dfrac{25}{36} = -\dfrac{1}{3} + \dfrac{25}{36} \)

\(x^2 + \dfrac{5}{3} x + \dfrac{25}{36} = -\dfrac{12}{36} + \dfrac{25}{36} \)

\(\left( x + \dfrac{5}{6} \right)^2 = \dfrac{13}{36} \)

Yes, that is what I got, too.

Answer 31

wow cool, thanks for such detailed explanations

Answer 32

The 3 is not added because it was not factored out. The 3 was divided by on both sides, thus eliminating the 3.

Answer 33

yeah in my book it says we have to multiply everything by 1/a

Answer 34

I'm impressed that you took your check all that distance and got what you'd hoped for!

Answer 35

thanks again both of you

so it also said like it had to be ax^2+bx = -c

what if c is not negative? do you have to multiply evrything by -1?

Answer 36

Yes, I agree with mathstudent55 about that 3 not being needed further.

Answer 37

In your shoes I'd just move that "c" to the other side of your equation.

Answer 38

Example: you have -2 on the right side; to get rid of that, add 2 to both sides of your equation. Result: 0 on the right side and left side has the form\[ax^2+bx+c\]

Answer 39

Here is a recap of the entire problem, since it was done in pieces throughout this post.

\(3x^{2}+5x+1 =0\)

\(3x^{2}+5x = -1\)

\(x^{2}+\dfrac{ 5 }{ 3 }x = -\dfrac{ 1 }{ 3 }\)

\(x^2 + \dfrac{5}{3} x + \dfrac{25}{36} = -\dfrac{1}{3} + \dfrac{25}{36}\)

\(x^2 + \dfrac{5}{3} x + \dfrac{25}{36} = -\dfrac{12}{36} + \dfrac{25}{36}\)

\(\left( x + \dfrac{5}{6} \right)^2 = \dfrac{13}{36}\)

All we did was complete the square.
If the goal is to solve the equation by completing the square, then you go on like this:

\( x + \dfrac{5}{6} = \pm \sqrt{\dfrac{13}{36}}\)

\( x + \dfrac{5}{6} = \pm \dfrac{\sqrt{13}}{6}\)

\( x = -\dfrac{5}{6} \pm \dfrac{\sqrt{13}}{6}\)

\(x = \dfrac{-5 + \sqrt{13}}{6} \) or \(x = \dfrac{-5 - \sqrt{13}}{6} \)

Answer 40

thanks!

Answer 41

You're welcome.