Question

if
$\alpha \: \: \: and \: \: \beta$
are the roots of
$8x {}^{2} - 6x + 3 = 0$
from an equation whose roots are
$\alpha - 3 \: and \: \beta - 3.$
​

Answer 1

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

Given that,

$\rm :\longmapsto\: \alpha \: and \: \beta \: are \: the \: roots \: of \: {8x}^{2} - 6x + 3 = 0$

We know,

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

$\rm :\longmapsto\: \alpha + \beta = - \dfrac{( - 6)}{8} = \dfrac{3}{4}$

and

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

$\rm :\longmapsto\: \alpha \beta = \dfrac{3}{8}$

Now, we have to form a quadratic equation whose roots

$\rm :\longmapsto\: \alpha - 3 \: and \: \beta - 3$

Let Consider,

Sum of zeroes,

$\rm :\longmapsto\: \alpha - 3 \: + \: \beta - 3$

$\rm \: = \: \:( \alpha + \beta ) - 6$

$\rm \: = \: \:\dfrac{3}{4} - 6$

$\rm \: = \: \:\dfrac{3 - 24}{4}$

$\rm \: = \: \: - \: \dfrac{21}{4}$

Now,

Consider, Product of zeroes,

$\rm :\longmapsto\: (\alpha - 3) \: \times \: (\beta - 3)$

$\rm \: = \: \: \alpha \beta - 3 \alpha - 3 \beta + 9$

$\rm \: = \: \: \alpha \beta - 3 (\alpha + \beta )+ 9$

$\rm \: = \: \:\dfrac{3}{8} - 3 \times \dfrac{3}{4} + 9$

$\rm \: = \: \:\dfrac{3}{8} - \dfrac{9}{4} + 9$

$\rm \: = \: \:\dfrac{3 - 18 + 72}{8}$

$\rm \: = \: \:\dfrac{57}{8}$

So, we have

$\boxed{\red{\sf Sum\ of\ the\ zeroes= - \: \frac{21}{4}}}$

and

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{57}{8}}}$

Now,

Required Quadratic equation is given by

$\rm :\longmapsto\: {x}^{2} - (Sum \: of \: zeroes)x + Product \: of \: zeroes = 0$

$\rm :\longmapsto\: {x}^{2} - {\bigg( - \dfrac{21}{4} \bigg) }x + {\bigg(\dfrac{57}{8} \bigg) } = 0$

$\rm :\longmapsto\: {x}^{2} + {\bigg( \dfrac{21}{4} \bigg) }x + {\bigg(\dfrac{57}{8} \bigg) } = 0$

$\rm :\longmapsto\: {8x}^{2} + 42x + 57 = 0$

Additional Information :-

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

$\boxed{ \sf{ \: \alpha + \beta + \gamma = - \frac{b}{a}}}$

$\boxed{ \sf{ \: \alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a}}}$

$\boxed{ \sf{ \: \alpha \beta\gamma = - \frac{d}{a}}}$