Question

Question #3c005

Answer 1

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.
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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`

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

Giving:`" "2x^2+3sqrt(2)x+2`
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This is how they look on the graph: