Question

The number of zeroes that polynomial f(x) = (x-2)^2 +4 can have is ​

Answer 1

0-2^2 =-4

-4+4 =0 is corrrect

Answer 2

$\bf \LARGE \color{pink}Hola!$

GiveN :

$\sf \rightsquigarrow \: Given \: \: polynomial \: \: f(x) = {(x - 2)}^{2} + 4$

To FinD :

$\sf \rightsquigarrow \: The \: \: number \: \: of \: \: Zeros$

$\sf \rightsquigarrow \: Zeros \: \: of \: \: the \: \: polynomial$

SolutioN :

~Expression :

$\: \: \sf \: f(x) = {(x - 2)}^{2} + 4$

→ For it to be zero,

$\implies \sf {(x - 2)}^{2} + 4 = 0$

$\implies \: \sf {x}^{2} - 4x + 4 + 4 = 0$

$\implies \: \sf {x}^{2} - 4x +8 = 0$

$\sf \: Sridharacharya \: \: formula, \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \tt \: x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} }$

$\implies \sf \: x = \frac{ - ( - 4) \pm \sqrt{ {( - 4)}^{2} - 4 \times 1 \times 8 } }{2 \times 1}$

$\implies \sf \: x = \frac{4 \pm \: \sqrt{ - 16} }{2}$

$\implies \sf \: x = {}{2} \: \pm \: i \: 4$

$\large \therefore \sf \: Hence, \: we \: can \: see \: there \: doesn't exist \: any \: \: real \: value \: for \:f( x )\: to\: be \: zero$

ConcepT BoosteR :

$\sf \rightsquigarrow \: Let's \: \: have \: \: a \: \: look \: \: in\: \: the \: \: graph \: \: of \: \: the \: \: polynomial$

$\underline{ \underline{ \mathbb {GRAPH \: \: IS \: \: IN \: \: THE \: \: ATTACHMENT}}}$

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HOPE THIS IS HELPFUL...

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