Which value of b will cause the quadratic equation x^2+bx+5=0 to have two real number solution

Every value of b>4.47 and b<-4.47 will cause the quadratic equation to have two real number solution.

Step-by-step explanation:

We have the quadratic function , and we have to find the value of b.

A quadratic function is , a quadratic function usually has two real solutions. You can find that solutions using Bhaskara's Formula:

and are real solutions of the quadratic equation if and only if:

If the quadratic equation doesn't have real solutions.

If    the quadratic equation has only one solution.

Then in this case to have two real number solutions:

We have , where a=1, b, c=5

Then,

Adding 20 in both sides of the equation:

Which is the same as:

Then,

Then every value of b>4.47 and b<-4.47 will cause the quadratic equation to have two real number solution.

For example b=-5 or b=5.

If you replace with b=-5 in

Then the quadratic function has two real number solutions.

given that triangle abc is transformed two times to get triangle

a"b"c"

first abc is transformed in to a'b'c' as follows:

the coordinates of a are (1,-1) and transformed into a'(-1,1)

similarly b (4,-2) became b'(-4,2) and

c(7,-2) became c'(-7,2)

i.e. (x,y) becomes (-x,-y)

this is nothing but reflection about a point here origin.

thus first transformation is reflection on the origin.

next is exactly shifting 3 units down

source:

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