Question #0429d

1 Answer
Feb 18, 2017

Answer:

#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#

Tony B