Help in math?

Question

3x^2 - x - 2 = 0

Help me solve this by completing the square

I need to show work. Please help me by showing work... I am in danger of failing...

Answer 1

Step 1: Get the constants all on one side:
3x^2 - x = 2
Step 2: Divide by the coefficient on x^2:

x^2 - (1/3) x = 2/3

Step 3: What is HALF of the coefficient on the x term?
-1/6

Step 4: Square the result from step 3:
1/36

Step 5: Add that to each side:
x^2 - 1/3 x + 1/36 = 2/3 + 1/36 (or 25 / 36 if we want to just add 'em together...)

---now, x + whatever_you_ wrote_down_in_step_3 is GUARANTEED TO BE, when you square it, the LEFT HAND SIDE of the equation. No, really. It is. Guaranteed. You follow these steps, man, and it IS.

So: (x - (1/6))^2 = x^2 - (1/3)x + 1/36
And we know that
(x-(1/6))^2 = 25/36, so
x-(1/6) = +/- 5/6 (square root of both sides)
x = either 5/6 + 1/6 or -5/6 + 1/6
or 6/6 or -4/6
1 or -2/3
Just remember the steps:
1. Move all the constant stuff to the right hand side.
2. Divide by the x^2 coefficient.
3. Write down what half of the x coefficient is (somewhere on the paper (sometimes this will be a negative number, so be careful).
4. Square the result of step 3 (this will ALWAYS be positive).
5. Add the result of step 4 to each side (adding up the fractions on the right side will VERY often prove to be a REALLY GOOD IDEA).
6. (x+whatever_you_wrote _down_in_step_3) is the square root of your perfect square! It JUST IS!
Bing! Boom! It's done!

(That's sorta how I teach it.)

Answer 2

Hi,

Completing the Square

3x^2 - x - 2 = 0

Move the constant to the other side.

3x^2 - x = 2

Since there is a coefficient in front of the squared term, divide everything by that number. Fractions are OK This time you need to do this.

3/3x^2 - 1/3x = 2/3
x^2 - 1/3x = 2/3

Multiply the fraction 1/2 times the number in the second term. Square your answer. Reduce if possible. Add to both sides.

(1/2)(-1/3) = -1/6

(-1/6)^2 = 1/36

Add 1/36 to both sides.

x^2 - 1/3x + 1/36= 2/3 + 1/36

Combine like terms. Then factor the left. It will always start with your variable, and have the number you squared above in your parentheses. The one sign in your factor should be the sign in front of the second term of your problem. This factor will always be squared. It can't be 2 different factors.

x^2 - 1/3x + 1/36= 2/3 + 1/36 becomes:
x^2 - 1/3x + 1/36= 25/36
(x - 1/6)^2 = 25/36

Now take the square root of both sides. That eliminates the parentheses and the power on the left, and it takes the square root on the right, with a PLUS or MINUS sign +/- in front of it.

(x - 1/6)^2 = 25/36
x - 1/6 = +/- 5/6

Now solve each of these equations:

x - 1/6 = 5/6
x = 1

x - 1/6 = -5/6
x = -2/3

Your 2 answers are 1 and -2/3

Here's another, so you can see the pattern.

a^2 - 4a - 21 = 0

Move the constant to the other side.

a^2 - 4a = 21

If there was a coefficient in front of the squared term, divide everything by that number. Fractions are OK This time you don't need to do this.

Multiply the fraction times the number in the second term. Square your answer. Reduce if possible. Add to both sides.

a^2 - 4a = 21

1/2 x -4 = -2 and (-2)^2 = 4, so add 4 to both sides.

a^2 - 4a + 4 = 21 + 4

Combine like terms. Then factor the left. It will always start with your variable, and have the number you squared above in your parentheses. The one sign in your factor should be the sign in front of the second term of your problem. This factor will always be squared. It can't be 2 different factors.

a^2 - 4a + 4 = 21 + 4 becomes:
a^2 - 4a + 4 = 25
(a - 2)^2 = 25

Now take the square root of both sides. That eliminates the parentheses and the power on the left, and it takes the square root on the right, with a PLUS or MINUS sign +/- in front of it.

(a - 2)^2 = 25
a - 2 = +/-5

Now solve each of these equations:

a - 2 = 5
a = 7

a - 2 = -5
a = -3

Your 2 answers are 7 and -3

I hope that helps you !! :-)

Answer 3

3x^2 - x - 2 = 0

Your first step would be to factor out 3 from the first two terms.

3(x^2 - (1/3)x) - 2 = 0

Now, add "half squared" of the coefficient of x inside of the brackets, and subtract 3 times that value outside of the brackets.

3(x^2 - (1/3)x + (1/36)) - 2 - 3(1/36) = 0

Factor the now-square trinomial.

3(x - (1/6))^2 - 2 - 3/36 = 0
3(x - (1/6))^2 - 2 - 1/12 = 0
3(x - (1/6))^2 - 24/12 - 1/12 = 0
3(x - (1/6))^2 - 25/12 = 0

3(x - (1/6))^2 = 25/12

Multiply both sides by (1/3),

(x - (1/6))^2 = 25/36

Here's the most important part; take the square root of both sides, _keeping in mind_ that we *have* to include plus-or-minus as part of the square root property.

x - (1/6) = +/- sqrt(25/36)

We have a fraction of perfect squares, so we can calculate its square root.

x - (1/6) = +/- 5/6

x = (1/6) +/- (5/6)

Therefore, the two values of x are

x = (1/6) + (5/6) , x = (1/6) - (5/6)

Simplified, they are

x = (6/6) = 1
x = (-4/6) = -2/3

Answer 4

First you have to factor the equation:

(x - 1)(3x + 2) = 0

Then, set each set equal to 0, because anything multiplied by the other must equal 0.

x-1=0 and 3x+2=0

So, solve each of these equations:

x=1
x=-2/3

Answer 5

x=4

Answer 6

3x^2 - 3x + 2x - 2 =0 (split the middle term)
3x(x - 1)+2(x - 1)=0
(3x+2)(x - 1)
x= -2/3 (or) 1




sHaKiRa

Answer 7

A= 3
B= -1
C= -2

DELTA = B*B - 4 * A * C
= 1 - 4 * 3 * (-2) = 1 +24 = 25 > 0

r1 = [ - B + sqrt(DELTA) ] / (2*A)
= 1

r2 = [ - B - sqrt(DELTA) ] / (2*A)
= -4/6 = -2/3

(x - r1) * (x - r2) = 0 -->
(x - 1) * (3x + 2) = 0 -->
3x^2 + 2x - 3x - 2 = 0 -->

3x^2 - x - 2 = 0

Answer 8

X = 1

3 times (1^2) = 3 right? then -1-2=0

There you have it.

Answer 9

Ok, you can always use:

opposite b plus or minus the square root of (b squared minus four (a)(c)) divided by 2a.

Answer 10

x=1
x=(-2/3)