How do you solve using the quadratic formula for #y = x^2 - 4x + 4#?

Question

Genesis Allen · Accepted Answer

The answer is : #y = x^(2) - 4x + 4 = (x-2)^(2)#. Your function has one zero, with #x = 2#. The usual form of  a quadratic function is : #y = ax^(2) + bx + c# Your function is #y = 1x^(2) - 4x + 4#.  Therefore #a = 1#, #b = -4# and #c = 4# The quadratic formula gives you the values of #x#, with which your #y = 0#. #x = (-b +- sqrtDelta)/(2a)#, where #Delta=b^(2) -4ac#.  Since we have a square root, #Delta >=0#. If not, the function don't have any zeros. Let's calculate the zeros of your function :  #Delta =(-4)^(2)-4*1*4=16-16=0# #x_1 = (4 +sqrt0)/2 = x_2 = (4-sqrt0)/2=#2 Therefore, #y = x^(2) - 4x + 4 = (x-2)^(2)#. Your function has one zero, with #x = 2#.

Nathan Axel · Answer

undefined To solve the equation $ y = x^2 - 4x + 4 $ using the quadratic formula, we first need to identify the coefficients $ a $, $ b $, and $ c $ in the general quadratic equation $ ax^2 + bx + c = 0 $. In this case:

$ a = 1 $
$ b = -4 $
$ c = 4 $

Next, we plug these values into the quadratic formula:

$$ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} $$

Substitute the values:

$$ x = \frac{{-(-4) \pm \sqrt{{(-4)^2 - 4(1)(4)}}}}{2(1)} $$

Simplify:

$$ x = \frac{{4 \pm \sqrt{{16 - 16}}}}{2} $$

$$ x = \frac{{4 \pm \sqrt{0}}}{2} $$

Since the discriminant $ b^2 - 4ac $ is zero, we have a repeated root. Therefore, the solutions are:

$$ x = \frac{4}{2} = 2 $$ (repeated root)