Question

Find the number of real zeros of P(x) = (x -2)^2 +3​

Answer 1

Answer:

0 real roots as (b^2-4ac) is less than 0

Answer 2

Question:

Find the number of real zeros of P(x) = (x -2)^2

Find:

Find the number of real zeros

Answer:

$x=2$

Step-by-step explanation:

$P(x) = (x -2)^2$

$P(x) = (x -2)^2 =0\\(x -2)^2 =0\\(x -2) =0\\x=2$

#SPJ2