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if `h(x) = x^2 – 5x – 11`. Solve `h(x) = 0`

if `h(x) = x^2 – 5x – 11`. Solve `h(x) = 0`

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if `h(x) = x^2 – 5x – 11`. Solve `h(x) = 0`

Postby barnamah » Wed 30 March, 2011 16:33

if `h(x) = x^2 – 5x – 11`. Solve `h(x) = 0`
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Re: if `h(x) = x^2 – 5x – 11`. Solve `h(x) = 0`

Postby Integr88 » Wed 30 March, 2011 16:50

to solve for `h(x) = x^2 – 5x – 11` we write it as ` x^2 – 5x – 11=0` and solve for x, therefore using completing the square we continue to factor it.

ad 11 to both side and we have `x^2 – 5x =11`

coefficient of the `5x` terms is 5. Divide it by 2 and square it `(-5/2)^2`.

Add `(-5/2)^2` to both side of equation and we have `x^2 – 5x+(-5/2)^2 =11+(-5/2)^2`

On the right side we have `x^2 – 5x+(-5/2)^2 =11+(25/4)` which is `x^2 – 5x+(-5/2)^2 =69/4`

Now on the left we have perfect square using `(-5/2)^2` and we can write it as `(x-5/2)^2=69/4`

`sqrt((x-5/2)^2)=+-sqrt(69/4)` and we `x-5/2=+-sqrt(69/4)`

add `5/2` to both side `x-5/2+5/2=+-sqrt(69/4)+5/2` and we have `x=+-sqrt(69/4)+5/2`

so x is
`x=-sqrt(69/4)+5/2=-1.653`
and
`x=+sqrt(69/4)+5/2=6.65`

`:.` x is `-1.653` and `x=6.65`
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