How do you solve #x^2-14x-49=0#?
↳Redirected from
"How do I change #int_0^1int_0^sqrt(1-x^2)int_sqrt(x^2+y^2)^sqrt(2-x^2-y^2)xydzdydx# to cylindrical or spherical coordinates?"
#x^2-14x-49=0#
This is unfactorable, therefore you would use the quadratic formula,
#x=(-b+-sqrt(b^2-4ac))/(2a)#
#a=1#
#b=-14 #
#c=-49#
Plug in the values a, b and c accordingly.
#x=(-b+-sqrt(b^2-4ac))/(2a)#
#x=(-(-14)+-sqrt((-14)^2-4(1)(-49)))/(2(1))#
#=(14+-sqrt(196+196))/(2)#
#=(14+-sqrt(392))/(2)#
#=(14+-14sqrt(2))/(2)#
#x=7+-7sqrt(2)#
#x=7+7sqrt2 or x=7-7sqrt2#
#x^2-14x - 49 =0#
Use the quadratic formula
#x=(-b+-sqrt(b^2-4ac))/(2a)#
Where #a=1, b=-14, c=-49#
#=(-(-14)+-sqrt((-14)^2-4(1)(-49)))/((2)(1)#
#x=(14+-sqrt(196+196))/(2)#
#x=(14+-sqrt392)/2#
#x= 7 + 7sqrt2 or x=7 - 7sqrt2#
The quadratic formula uses a quadratic equation. The equation looks like this:
#ax^2+bx+c#
...and the formula looks like this:
#x=(-b+-sqrt(b^2-4ac))/(2a)#
For this setup:
#a=1#
#b=-14#
#c=-49#
Plugging that into the formula:
#x=(-(-14)+-sqrt((-14)^2-4(1)(-49)))/(2(1))#
#x=(14+-sqrt(196+196))/(2)#
#x=(14+-sqrt(2xx196))/(2) rArr x=(14+-14sqrt(2))/(2)#
#x=7+-7sqrt(2)rArr color(red)(x={16.8995,-2.8995}#
