Question

21. Ifa and ß are zeroes of the quadratic polynomial x2 - 6x + a the value of 'd', if 30 + 2B = 20 is - 12 (a) True (b) False (c) Can't say (d) Partially True/False​

Answer 1

Answer:

it is false

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Answer 2

Appropriate Question :-

If α and β are zeroes of the quadratic polynomial x^2 - 6x + a, then the value of a, if 3α + 2β = 20 is - 12 is

(a) True

(b) False

(c) Can't say

(d) Partially True/False

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

Given that

$\rm :\longmapsto\: \alpha, \beta \: are \: zeroes \: of \: {x}^{2} - 6x + a$

We know,

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\rm\implies \: \alpha + \beta = - \dfrac{( - 6)}{1}$

$\rm\implies \: \alpha + \beta = 6$

$\rm\implies \: \beta = 6 - \alpha - - - - (1)$

Also, given that

$\rm :\longmapsto\:3 \alpha + 2 \beta = 20$

$\rm :\longmapsto\:3 \alpha + 2 (6 - \alpha ) = 20$

$\red{ \bigg\{ \sf \: \because \: using \: equation \: (1) \bigg\}}$

$\rm :\longmapsto\:3 \alpha + 12 - 2 \alpha = 20$

$\rm :\longmapsto\: \alpha = 20 - 12$

$\rm\implies \:\boxed{\tt{ \: \: \alpha = 8 \: \: }} - - - (2)$

On substituting equation (2) in (1), we get

$\rm :\longmapsto\: \beta = 6 - 8$

$\rm\implies \:\boxed{\tt{ \: \: \beta = - 2 \: \: }} - - - - (3)$

Also, We know that

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\rm\implies \: \alpha \beta = \dfrac{a}{1}$

$\rm\implies \: \alpha \beta = a$

$\bf\implies \:a = 8( - 2) = - 16$

$\rm\implies \:\boxed{\tt{ \: \: a \: = \: - \: 16 \: \: }}$

So, option (b) is Correct

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Explore More

For Cubic Polynomial

$\red{\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: zeroes \: of \: a {x}^{3} + b {x}^{2} + cx + d, \: then}$

$\boxed{ \bf{ \: \alpha + \beta + \gamma = - \dfrac{b}{a}}}$

$\boxed{ \bf{ \: \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}}}$

$\boxed{ \bf{ \: \alpha \beta \gamma = - \dfrac{d}{a}}}$