Question #3c005

1 Answer
Aug 11, 2016

Answer:

There are 2 Real roots

Explanation:

Consider the standard form equation: #y=ax^2+bx+c#

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

Now consider the discriminate: #b^2-4ac=10#

The fact that this is positive means that the solution for #x# belongs to the set of Real Numbers: #x in RR#

Thus we have #(-b)/(2a)+-(sqrt(10))/(2a)#

Also #sqrt(10)/(2a) !=0 and (-b)/(2a) !=0#

Let #x_1=(-b)/(2a)+(sqrt(10))/(2a) #

Let #x_2=(-b)/(2a)-(sqrt(10))/(2a)#

Thus #x_1-x_2 =cancel(-(b)/(2a))+(sqrt(10))/(2a)cancel(+ (b)/(2a))+(sqrt(10))/(2a)#

#x_1-x_2" "=" " cancel(2)xxsqrt(10)/(cancel(2)a)#

As there is a defined distance between the two points then they do not coincide. Thus there are two separate points.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Lets try out two equations #" "b^2-4ac# by making up some numbers.

Suppose we had:#" "b^2-4(2)(10) =10#
#=>b^2=10+80=90" "->" "b=3sqrt(10)#

Giving:#" "2x^2+3sqrt(10)x+10#

'.........................................................................
Suppose we had:#" "b^2-4(1)(2)=10#
#=>b^2=10+8" "->" "b=3sqrt(2)#

Giving:#" "2x^2+3sqrt(2)x+2#
'....................................................................
This is how they look on the graph:
Tony B