Solve b²-4b+4=0 using the quadratic formula explain has test on

b = 2

Step-by-step explanation by completing the square:

Solve for b over the real numbers:

b^2 - 4 b + 4 = 0

Write the left hand side as a square:

(b - 2)^2 = 0

Take the square root of both sides:

b - 2 = 0

Add 2 to both sides:

 b = 2

I'm going to use x in place of b. The reason why is because 'b' shows up in the quadratic formula, so there might be some confusion.

So we want to solve x^2 - 4x + 4 = 0 for x

Note how this equation is the same as 1x^2 - 4x + 4 = 0 and it is in the format ax^2 + bx + c = 0

We see that a = 1, b = -4 and c = 4

We will plug these values into the discriminant formula

d = b^2 - 4ac

d = (-4)^2 - 4(1)(4)

d = 16 - 16

d = 0

The discriminant is 0, so the square root of this is also 0. A discriminant of 0 means that there is only one real solution.

It turns out that the solution is equal to x = -b/(2*a) = -(-4)/(2*1) = 4/2 = 2

If the discriminant were nonzero, then the quadratic formula would be a bit messier.

It might be better to factor. Doing so gets you (x-2)^2 = 0 which becomes x-2 = 0 and x = 2

step-by-step explanation:

here's what happens:

1) the graph of f   = x^2 is shifted 2 units to the right.

2) the resulting graph is reflected in the x-axis, so that the parabola now opens down.

3) this final result is transformed up by 5 units.

need more info

step-by-step explanation: