# If the zeroes of the polynomial $x^3 - 3x^2 + x + 1$ are $a-b, a, a + b$, find $a$ and $b$.

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Given:

The zeroes of the polynomial $x^3 - 3x^2 + x + 1$ are $a-b, a, a + b$.

To do:

We have to find $a$ and $b$.

Solution:

Let $\alpha, \beta$ and $\gamma$ be the zeroes of polynomial $x^3 – 3x^2 + x + 1$.

This implies,

$\alpha = a-b, \beta = a$ and $\gamma = a + b$.

Therefore,

Sum of zeroes $= \alpha + \beta + \gamma$

$(a – b) + a + (a + b)=3$

$a-b + a + a + b = 3$

$3a = 3$

$a= 1$........…(i)

Product of zeroes $= \alpha \beta \gamma$

$(a – b) a (a + b) = -1$

$(a^2 – b^2)a = -1$

$a^3 – ab^2 = -1$...… (ii)

Putting the value of $a$ from equation (i) in equation (ii), we have,

$(1)^3-(1)b^2 = -1$

$1 – b^2 = -1$

$b^2 = 1 + 1$

$b^2 = 2$

$b = \pm \sqrt2$

Hence, $a = 1$ and $b = \pm \sqrt2$

Updated on 10-Oct-2022 13:19:38