Determine the number of real solutions y= x^2-4x+4? A. 0 B. 1 C. 2 D. 3

#1# real solution

Explanation:

The key realization is that when we think about how many solutions a quadratic of the form

#ax^2+bx+c#

will have, we like to look at the discriminant, #b^2-4ac#. This is notoriously the quantity under the radical in the Quadratic Formula.

We have three cases:

#b^2-4ac>0=>2# real solutins

#b^2-4ac<0=># no real solutions

#b^2-4ac=0=>1# real solution

In our quadratic, we know #a=1, b=-4, c=4#, so we can plug these values into our expression for the discriminant:

#16-4(4)=16-16=0#

We see that our discriminant is zero, which means this quadratic has one real solution.

Hope this helps!