Question

if a and b are the zeroes of equation 2x^2+4x+5 find the value of a^2+b^2​

Answer 1

➤Question :-

$\rm if \: a \: and \: b \: are \: the \: zeros \: of \: equation \: \\ \rm 2 {x}^{2} + 4x + 5 \: then \: find \: the \: value \: of \: \\ \rm {a}^{2} + {b}^{2} .$

➤Answer :-

$\mapsto \boxed{\rm {a}^{2} + {b}^{2} = - 1 \: } \:$

➤To Find :-

$\mapsto \rm {a}^{2} + {b}^{2}$

➤Used Formula :-

$\to \rm {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab$

➤Solution :-

According to the question,

$\rm a \: and \: b \: are \: the \: zeros \: of \: equation \: \\ \rm 2 {x}^{2} + 4x + 5 \: \: \\ \bf therefore \\ \\ \because \star \rm sum \: of \: zeros = \frac{ - coefficient \: of \: x}{coefficient \: of \: {x}^{2} } \\ \\ \mapsto \boxed{ \rm a + b = \frac{ - 4}{2} = - 2} \: \: \: \: .....eq.(1) \\ \\ \bf and \: \\ \because \star \rm product \: of \: zero = \frac{ constant \: term}{coefficient} \\ \\ \mapsto \boxed{ \rm \: ab = \frac{5}{2} } \: \: \: \: ....eq.(2) \\ \\ \sf hence \\ \\ \mapsto \rm {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \\ \\ \mapsto \rm {a}^{2} + {b}^{2} = {(a + b)}^{2} - 2ab \\ \\ \rm \: from \: eq.(1) \: and \: eq.(2) \\ \\ \mapsto \rm {a}^{2} + {b}^{2} = {( - 2)}^{2} - 2 \times \frac{5}{2} \\ \\ \mapsto \rm {a}^{2} + {b}^{2} = 4 - 5 \\ \\ \mapsto \boxed{\rm {a}^{2} + {b}^{2} = - 1 \: }$