Question

if the zeroes 9f the polynomial x³-3x²+x+1 are a-b,a,a+b find a and b​

Answer 1

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

We know that

$\sf\:If\:\alpha,\beta,\gamma\:are\: zeroes \: of \: f(x)={ax}^{3}+ {bx}^{2}+cx +d\:then$

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{3}}}}$

Or

$\boxed{\purple{\tt Sum\ of\ the\ zeroes, \alpha+\beta + \gamma =\frac{-b}{a}}}$

And

$\boxed{\red{\sf Product\ of\ the\ zeroes= - \dfrac{Constant}{coefficient\ of\ x^{3}}}}$

Or

$\boxed{\purple{\tt Product\ of\ the\ zeroes, \: \alpha\beta\gamma = - \: \frac{d}{a}}}$

Let's solve the problem now!!

Given that

$\sf\:a - b, \: a, \: a + b\:are \: the\: zeroes \: of \:$

$\sf \: f(x)={x}^{3} - 3 {x}^{2}+x +1$

So,

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{3}}}}$

$\rm :\longmapsto\:a - \cancel{b} + a + a + \cancel{b} = - \dfrac{( - 3)}{1}$

$\rm :\longmapsto\:3a = 3$

$\bf :\implies\:a \: = \: 1$

Now,

We know,

$\boxed{\red{\sf Product\ of\ the\ zeroes= - \dfrac{Constant}{coefficient\ of\ x^{3}}}}$

$\rm :\longmapsto\:(a - b)a(a + b) = - \dfrac{1}{1}$

$\rm :\longmapsto\:(1 - b)(1 + b) = - 1 \: \: \: \{ \: as \: a \: = \: 1 \}$

$\rm :\longmapsto\:1 - {b}^{2} = - 1$

$\rm :\longmapsto\: {b}^{2} = 2$

$\bf\implies \:b = \pm \: \sqrt{2}$

Additional Information :-

$\sf\:If\:\alpha,\beta\:are\: zeroes \: of \: f(x)={ax}^{2}+ {bx}+c\:then$

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

Or

$\boxed{\purple{\tt Sum\ of\ the\ zeroes, \: \alpha + \beta =\frac{-b}{a}}}$

And

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

Or

$\boxed{\purple{\tt Product\ of\ the\ zeroes, \: \alpha \beta =\frac{c}{a}}}$