Question

using graph find zeroes of y=x^2-4x+5​

Answer 1

Step-by-step explanation:

x= 1,5

thanku ...................

Answer 2

Required Answer :

• From this graph we can see there is no real value of x for which f ( x ) = 0 or we can say
• $\tt \: x \: \: \cancel\in \: \: \R$

~Verification

$f(x) = {x}^{2} - 4x + 5 \\ \\ \\ \underline{ \underline{for \: \: it \: \: to \: \: be \: \: real \: \: root }} \\ \\ D \geqslant 0 \\ \\ : \to\sqrt{ {( - 4)}^{2} - 4 \times 1 \times 5 } \geqslant 0 \\ \\ : \to \: 16 - 20 \geqslant 0 \\ \\ \bf \: which \: \: is \: \: not \: \: possible \\ \\ \\ \therefore \: \bf \: roots \: \: are \: \: not \: \: on \: \: the \: \: real \: \: axis$

Hence Verified!!