Find the number of real zeros of P(x) = (x - 2)^2+3:(1) 1(2) 0(3) 2(4)

1.	Find the number of real zeros of P(x) = (x - 2)^2+3:(1) 1(2) 0(3) 2(4) None of thesePlease answer with solution
Answer» ANSWER: OPTION (2). 0 Step-by-step EXPLANATION: $p(x) = {x}^{2} - 4x + 4 + 3 \\ \\ = {x}^{2} - 4x + 7 \\ \\ d = {b}^{2} - 4ac \\ \\ = {( - 4)}^{2} - 4(1)(7) \\ \\ = 16 - 28 \\ \\ = - 12 < 0 \\ \: \: \: \: \: \: \\ \\ \\ d < 0\\ \\ option \: (2) \: \: 0 \: \: \: \: \: \: \: \: is \: \: \: correct. \\ \\$

Find the number of real zeros of P(x) = (x - 2)^2+3:(1) 1(2) 0(3) 2(4) None of thesePlease answer with solution

Answer»

Step-by-step EXPLANATION:

$p(x) = {x}^{2} - 4x + 4 + 3 \\ \\ = {x}^{2} - 4x + 7 \\ \\ d = {b}^{2} - 4ac \\ \\ = {( - 4)}^{2} - 4(1)(7) \\ \\ = 16 - 28 \\ \\ = - 12 < 0 \\ \: \: \: \: \: \: \\ \\ \\ d < 0\\ \\ option \: (2) \: \: 0 \: \: \: \: \: \: \: \: is \: \: \: correct. \\ \\$

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Find the number of real zeros of P(x) = (x - 2)^2+3:(1) 1(2) 0(3) 2(4) None of thesePlease answer with solution​

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Find the number of real zeros of P(x) = (x - 2)^2+3:(1) 1(2) 0(3) 2(4) None of thesePlease answer with solution