Question

if α,β are the zeroes of x^2 -6x+k what is the value of k.if 3α +2β =20?

Answer 1

$GIVEN P(x) = x^2 - 6x + k ..... (1) \\ 3α +2β =20 .............. (2)\\ MAKE polynomial zero (1)\\ x^2 - 6X + k = 0 \\ as \: per \: polynomials \: \\ a = 1 \\ b = - 6 \\ c = k \\ \alpha + \beta = - \frac{b}{a} \\ \alpha + \beta = {6} .....(3) \\ product \: \alpha \beta = \frac{c}{a} \\ \alpha \beta = k \\ by \: (2)and(3) \\ equate by \: multiply \: (3) \: by \: 3 \\ 3 \alpha + 3 \beta = 18 .....(4)\\ sub \: (4)from \: (2) \\ (3 \alpha - 3 \alpha ) + (2 \beta - 3 \beta ) = 20 - 18 \\ - \beta = 2 \\ \beta = - 2 \\ substitute \: \beta \: value \: in \: (4) \\ 3 \alpha + 3( -2 ) = 18 \\ 3 \alpha - 6 = 18 \\ 3 \alpha = 18 + 6 \\ 3 \alpha = 24 \\ \alpha = 8 \\ now \: place \: the \: value \: of \: \alpha \: and \: \beta \: in \: \alpha \beta = k \\ therefore \: 8 \times - 2 = k \\ k = - 16$

Answer 2

$\huge \underbrace{\boxed{ \bf\dag \green{ Given} \dag }}$

$\tt \alpha , \beta \: are \: zeroes \: of \: {x}^{2} - 6x + k$

$\tt3 \alpha + 2 \beta = 20$

$\huge \underbrace{\boxed{ \bf\dag \red{ FIND} \dag }}$

$\tt y = ?$

$\huge \underbrace{\boxed{ \bf\dag \pink{ Solution} \dag }}$

$\tt {x}^{2} - 6x + y \\ \tt \alpha + \beta = \frac{ - b}{a} = \frac{ \cancel - ( \cancel - 6)}{1} = 6.....(i) \\ \tt \alpha \beta = \frac{c}{a} = \frac{a}{1} = k .....(ii) \\ \tt multipy \: eq(i) \: by3 \\ \tt 3( \alpha + \beta ) = 3(6) \\ \tt 3 \alpha + 3 \beta = 18.....(iii) \\ \tt given \: eq. \: is, \: 3 \alpha + 2 \beta = 20.....(iv) \\ \tt subtracting \: eq(iv)from(iii), \\ \tt 3 \alpha + 3 \beta = 18 \\ \tt \underline{ - 3 \alpha + 2 \beta = 20} \\ \tt \underline{ \: \: \: \: \beta \: \: = \: \: \: \: \: \: \: - 2} \\ \tt from \: eq(i), - 2 + \beta = 6 \\ \tt \implies \beta = 6 + 2 = 8 \\ \tt now, \: from \: eq(ii), \\ \tt \alpha \beta = k \\ \tt \therefore k = ( - 2)(8) = - 16$

$\underbrace{\boxed{ \bf \purple{ Answer}}}$

$\boxed{ \tt value \: of \: k \: is - 16.}$