Question

What are all the real zeros of y=(x-20)^2 +5 Please show step by step I'm so lost!!

Answer 1

Answer:

No real zeros for the function.

Explanation:

• zero of a function f, if f(r) = 0

solve:

$\sf y=(x-20)^2 +5$

so when f(y) = 0

$\hookrightarrow \sf (x-20)^2 +5 = 0$

$\hookrightarrow \sf (x-20)^2 = -5$

$\hookrightarrow \sf (x-20) = \sqrt{-5}$

$\hookrightarrow \sf x = \pm \sqrt{-5} +20$

$\hookrightarrow \sf x = \sqrt{-5} +20, \ \sf x = -\sqrt{-5} +20$

check:

$\hookrightarrow f \sf (\sqrt{-5} +20) = Not \ applicable$, $f \sf (-\sqrt{-5} +20) = Not \ applicable$

Answer 2

Answer:

No real zeros.

Step-by-step explanation:

Given... y = $(x-20)^{2} +5$

Since above equation has no real real value, there are no zeros.

But complex zeros are:

         $(x-20)^{2} +5=0$

          $(x-20)^{2} =-5$

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