using completing the square solve `3x^2+10x-8=0`

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using completing the square solve `3x^2+10x-8=0`

Postby barnamah » Fri 18 March, 2011 06:41

Completing the Square
sometimes we get quadratic equations that cannot be solved by factoring. One of the method of finding the value of x is completing the square.

The questions says:
using completing the square solve `3x^2+10x-8=0`

in this equation the problem is with the coefficient of the squared term `x^2`. The 3 infront of `x^2`. That 3 is the determining factor of this equation. Here are the steps:

1- Divide every term by 3
`3/3x^2+10/3x-8/3=0` the first term cancels out the 3 and our equation becomes:
`x^2+10/3x-8/3=0`

2- Move the term without x (the `-8/3` term) to the right of the equation we have:
`x^2+10/3x=8/3`
The reason we moved the `-8/3` to the right is that because we have `-` attached to `8/3` , it is called negative term so to get rid of it, we add equal positive term to both side of equation. Here we have `-8/3`, so we add `+8/3` to both side of equation equation and we have:
`x^2+10/3x-8/3+8/3=+8/3`. The `-8/3+8/3` on the left cancel out each other and our equation becomes `x^2+10/3x=8/3`


3- Now on the left we have `x^2+10/3x`. The first term `x^2` which is second degree and the second term `10/3x` which is first degree. The coefficient of x is `10/3`.
Divide `10/3` by 2 (remember dividing by 2 is the same as multiplying by `1/2`):
`10/3 * 1/2=5/3`

4- Square the `5/3` and it becomes `(5/3)^2` which becomes `25/9` and add it to both side of the equation:
`x^2+10/3x+25/9=8/3+25/9`
simplify the right side and we have:
`x^2+10/3x+25/9=49/9`

5-Now the payoff time. Work we have done so far was to simply be able to factor the equation. Now the equation it is a perfect square:
`(x+5/3)^2=49/9`
Remember our aim is to factor the equation! which means find value(s) for `x`


6- Take the square root of both side of the equation:
`sqrt((x+5/3)^2)=sqrt((49/9))` and we have
`x+5/3= +-7/3`
what does `+-` mean?
it means we have either `7/3` or `-7/3`


7- Isolate the x by moving `5/3` to the right by changing its sign
`x=-5/3 +-7/3` so we have:
`x=-5/3 -7/3=2/3` so we have `x=2/3` and
`x=-5/3 +7/3=-4` `x=-4`

Now we equate them by `0` so for `x=-2/3` we multiply both side by `3*x=-2/3*3`. We will have on the right side `3*x=-2` and now move `-2` to the left by changing its sign to + and we will have `(3x+2)=0`.

Do the same for `x=-4` and move `-4` to the right and we have `(x+4)=0`

Putting them together we have `(3x+2)(x+4)=0`

`:.3x^2+10x-8=(3x+2)(x+4)=0`
;)
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barnamah
 
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