Question

if lemma and beta are the zeroes of p(x)=x²-5x+6, find the value of lemma +beta -3lemma×beta

Answer 1

Question :-

If lambda and beta are the zeros of p(x) = x² - 5x + 6 , Find the value of $\sf{ \lambda + \beta - 3 \lambda\:\beta }$

Answer :-

$\red{ \boxed{ \leadsto}} \boxed{\lambda + \beta - 3 \lambda\:\beta \: = - 13 \: }$

Solution :-

Given that :-

$\sf{\lambda \: \: and \: \: \beta \: are \: the \: zeroes \: of }\: \: \\ \to \sf{ \: p(x) = {x}^{2} - 5x + 6}$

We know that ;

$\boxed{ \sf{Sum \: of \: zeroes \: = - \frac{coefficient \: of \: x}{coefficient \: of \: {x}^{2} } }} \\ \\ \boxed{\sf{ Product \: of \: zeroes = \frac{constant \: term}{coefficient \: of \: {x}^{2} } }}$

Hence ,

$\star \: \bf{Sum \: of \: zeroes \: }\: \\ \red{ \implies} \boxed{ \lambda + \beta = \frac{5}{1} } \\ \\ \star \: \bf{Product \: of \: zeroes } \: \\ \green{\implies \: } \boxed{\lambda \: \beta = \frac{6}{1} }$

Therefore ,

$\pink{ \implies} \: (\lambda + \beta )- 3 (\lambda\:\beta) \: \\ \\ \implies \: \frac{5}{1} - 3 \times \frac{6}{1} \\ \\ \blue{ \implies \: }5 - 18 = - 13 \:$