Question

Question #0429d

Answer 1

`x=5+-3sqrt(3)`

Explanation:

Really worth committing this bit to memory if you can.

Consider the standardised formula: `" "y=ax^2+bx+c`

Where: `" "x=(-b+-sqrt(b^2-4ac))/(2a)`

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Given:`" "x^2-10x-2=0`

Thus: `a=1"; "b=-10"; "c=-2` giving:

`x=(-(-10)+-sqrt((-10)^2-4(1)(-2)))/(2(1))`

`x=(+10)/2+-(sqrt(100+8))/2`

`x=5+-sqrt(108/4)`

`x=5+-sqrt(27)`

`x=5+-sqrt(3xx3^2)`

`x=5+-3sqrt(3)`
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This is a sort of cheat method to determine the vertex x-value

Write in the form: `a(x^2+b/ax)+c`

Then `x_("vertex")=(-1/2)xxb/a`

In this case `a=1` as in `x^2->1x^2` giving:

`x_("vertex")=(-1/2)xxb/1" "->(-1/2)xx(-10) = +5`