Question

if a and b are given zeros of x ² - 6x + k find the value of k if 3a + 2b = 20
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Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:a \: and \: b \: are \: zeroes \: of \: polynomial \: {x}^{2} - 6x + k$

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 :-

$\boxed{ \sf \: { \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta )}^{2} - 2 \alpha \beta }$

$\boxed{ \sf \: { \alpha }^{3} + { \beta }^{3} = {( \alpha + \beta )}^{3} - 3 \alpha \beta( \alpha + \beta ) }$

$\boxed{ \sf \: \alpha - \beta = \sqrt{ {( \alpha + \beta) }^{2} - 4 \alpha \beta } }$