Question

If α, β are zeroes of x2 –6x + k and 3α + 2β = 20, find α, β.

Answer 1

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

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

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

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

$\tt y = ?$

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

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

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

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