Confused on finding the values of a variabie in a function with multiple variables.

[Chaim]

Well the equation is x^2 + bx + 3b
So I'm having trouble with this, cause the last thing has a variable
I'm suppose to find the value of b
Vertex is on the x-axis

This is what I did to start off with f(x) = x^2 + bx + 3b
When vertex is on the x-axis, y=0
(-b/2a) = (-b/2) = x
0=(-b/2)^2 + b(-b/2)+3b
0=(b^2/4) + -b^2/2 + 3b
Then I multipled everything by 4
0=b^2 - (4b^2/2) + 12b = 0
Then I got lost after this part, cause the 'b' part is weird so it's tough to do the quadratic equation on this.

Then after that is f(x) = x^2 + 3bx - b^2 + 1
So may anyone help me try to figure out what am I suppose to do next?
Thanks

 

[Deleted member 4993]

Well the equation is x^2 + bx + 3b
So I'm having trouble with this, cause the last thing has a variable
I'm suppose to find the value of b
Vertex is on the x-axis

This is what I did to start off with f(x) = x^2 + bx + 3b
When vertex is on the x-axis, y=0
(-b/2a) = (-b/2) = x
0=(-b/2)^2 + b(-b/2)+3b
0=(b^2/4) + -b^2/2 + 3b
Then I multipled everything by 4
0=b^2 - (4b^2/2) + 12b = 0
Then I got lost after this part, cause the 'b' part is weird so it's tough to do the quadratic equation on this.

Then after that is f(x) = x^2 + 3bx - b^2 + 1
So may anyone help me try to figure out what am I suppose to do next?
Thanks

If the vertex is on the x-axis → the discriminant is equal to zero.

b² - 12b = 0 → b(b-12) = 0 → b=0 or b=12

 

[Chaim]

If the vertex is on the x-axis → the discriminant is equal to zero.

b² - 12b = 0 → b(b-12) = 0 → b=0 or b=12

Ok thanks
But can you check the steps and see what I should do to get those answers
Cause my teacher didn't really teach us the discriminant way, so I was wondering if you could help me with the steps

 

[Deleted member 4993]

Well the equation is x^2 + bx + 3b
So I'm having trouble with this, cause the last thing has a variable
I'm suppose to find the value of b
Vertex is on the x-axis

This is what I did to start off with f(x) = x^2 + bx + 3b
When vertex is on the x-axis, y=0
(-b/2a) = (-b/2) = x
0=(-b/2)^2 + b(-b/2)+3b
0=(b^2/4) + -b^2/2 + 3b

0 = -b^2/4 + 3b

12b - b^2 = 0

b(12 - b) = 0 → b = 0 or b = 12

Then I multipled everything by 4
0=b^2 - (4b^2/2) + 12b = 0
Then I got lost after this part, cause the 'b' part is weird so it's tough to do the quadratic equation on this.

Then after that is f(x) = x^2 + 3bx - b^2 + 1
So may anyone help me try to figure out what am I suppose to do next?
Thanks

.

 

[Chaim]

If you have to use vertex here then tell me what is the co-ordinate of the vertex when the equation of the parabola is:

y = Ax² + Bx + C

Well I did (-b/2a) = -b/2(1) = -b/2 = x
So x = -b/2
The y, I think it is 0 because since the vertex is on the 'x' axis
So the vertex = (-b/2, 0)