There are 5 quadratics below. Four of them have two distinct roots each. The other has only one distinct root; find the value of that root.
x^2 - 4x + 5
9x^2 - 6x + 4
x^2 + 26x + 169
x^2 - 7x + 49
4x^2 + 4x + 9
+421 Do the discriminant; b^2-4ac, we must have b^2 -4ac = 0: first one: 16-20 ≠0, second one: 36 - 4*36 ≠0, third one: 26^2 - 4 * 13^2 ... yay! That equals 0. Another way we could do this is we could factor each and see which one is a perfect square binomial. So it is x^2 + 26x + 169, which factors as (x+13)^2, so -13, or vieta's formulas on -b/a to get -26/1 is the sum of the roots, except we double count so it is -13 either way!
