If a and b are given zeros of x ² - 6x + k find the value of k if 3a

1.	If a and b are given zeros of x ² - 6x + k find the value of k if 3a + 2b = 20
Answer» $\large\underline{\sf{Solution-}}$ Given that, $\rm :\longmapsto\:a \: and \: <klux>B</klux> \: are \: zeroes \: of \: polynomial \: {x}^{<klux>2</klux>} - 6x + k$ We know that, $\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$ $\bf\implies \:a + b = - \dfrac{( - 6)}{1} = 6 - - (1)$ Also, $\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$ $\bf\implies \: ab= \dfrac{k}{1} = k - - (2)$ Now, Given that, $\rm :\longmapsto\:3a + 2b = 20$ $\rm :\longmapsto\:a + 2a + 2b = 20$ $\rm :\longmapsto\:a + 2(a + b) = 20$ $\rm :\longmapsto\:a + 2 \times 6 = 20 \: \: \: \: \: \: \: \{using \: (1) \}$ $\rm :\longmapsto\:a + 12 = 20$ $\bf\implies \:a = 8 - - - (3)$ On substituting the value of a in EQUATION (1), we GET $\rm :\longmapsto\:8 + b = 6$ $\bf\implies \:b = - 2 - - - (4)$ Now, On substituting the values of a and b in equation (2), we get $\rm :\longmapsto\:k = 8 \times ( - 2)$ $\bf\implies \:k = - 16$ Additional Information :- $\boxed{ \sf \: { \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta )}^{2} - 2 \alpha \beta }$ $\boxed{ \sf \: { \alpha }^{3} + { \beta }^{3} = {( \alpha + \beta )}^{3} - 3 \alpha \beta( \alpha + \beta ) }$ $\boxed{ \sf \: \alpha - \beta = \sqrt{ {( \alpha + \beta) }^{2} - 4 \alpha \beta } }$