Need help solving this quadratic equation using the zero-factor property?

You're on the right track.

8/3*a^2 = 1 - 10/3a

You want to multiply both sides by 3. As long as you apply operations to both sides of the equation, you do not change the equation.

3*(8/3*a^2) = 3*(1-10/3*a)

8a^2 = 3 - 10a

(don't forget that 3*8/3, the 3's cancel. You still ended up with 3 times 8a. Also remember that it is a^2 and not just a)

Now you want to bring all the terms to one side and set it equal to zero

8a^2 + 10a - 3 = 0

And now you can factor. I'm assuming you know how to factor, but in case you do not I will point you to this tutorial:

http://www.purplemath.com/modules/factquad.htm (in this case you'll want to factor the hard case)

If you do factor you will find the answer is:

(2a+3)(4a -1) = 0

Now either 2a+3 = 0 or 4a-1 =0 since if either one of the terms is zero, and you multiply it by the other term, you still get zero.

a = -3/2 or a = 1/4

If you substitute this into the original equation you will find they do indeed equal zero. These are the roots of your equation.

Hope this helps

Cheers

Multiply by three and get rid of the fractions:

3[ 8/3 a^2 = 1 - 10/3a]

8a²=3 - 10a

Move the 10a and the 3 over.

8a² + 10a - 3 = 0

Now this is a quadratic equation that you can factor by guess and check or by grouping. I prefer factoring by grouping. To factor by grouping, multiply the coefficient of the squared term by the constant at the end, in this case, 8*-3 = -24. Are there factors of -24 that add to the middle 10? Yes, there are, +12, and -2.

Rewrite the equation with these two numbers replacing the +10a.

8a² + 12a - 2a - 3 = 0

Factor the equation by grouping the first two terms and the last two terms.

(8a² + 12a) +( - 2a - 3) = 0

Factor the common factor from each:

4a(2a + 3) -1(2a + 3) = 0

Now refactor the common (2a + 3)

(2a + 3)(4a - 1) = 0

Solve each one:

2a + 3 = 0

2a = -3

a = -3/2

4a - 1 = 0

4a = 1

a =1/4

These are your two answers.

8/3 a² = 1 - 10/3 a

24/3 a² = 3 - 30/10 a....(Multiply each term by 3, the LCD)

8a² = 3 - 10a

8a² + 10a - 3 = 0

(2a + 3)(4a - 1) = 0

If the product of two terms equals zero, then one or both terms equal zero.

If 2a + 3 = 0,

2a = - 3

a = - 3/2

If 4a - 1 = 0,

4a = 1

a =1/4

a {- 3/2, 1/4}

¯¯¯¯¯¯¯¯¯¯¯¯

1 = 3/3

8/3 a^2 = 1 - 10/3 a

8/3 a^2 = 3/3 - 10/3 a

1/3 (8 a^2) = 1/3 (3 - 10 a)

8a^2 = 3 - 10a

8a^2 + 10a -3 = 0

(4a -1)(2a+3) = 0

that's very ordinary, only look on the equation and look at what 2 formula you multiply to get the unique. n^2 + 9n + 18 =0 ??? initiate by using using writing (n + _)(n + _) and look at what plus what = 9, and what situations what = 18. those 2 numbers could be waiting to do the two. racking your recommendations you get (n + 3)*(n + 6) = 0 to income that that's right, only do the multiplication returned: n^2 + 6n + 3n +18 = n^2 + 9n + 18. confident that's right! the zeros can now be examine off as n = -3, -6