Question

if the zeroes of the polynomial x3-3x2+x+1are a/r, a, ar then the value of a is​

Answer 1

Step-by-step explanation:

Clearly, sum of the roots is,

a+(a-b)+(a+b)= -(-3)/1=3

=> 3a= 3

=> a= 1.

Again, product of roots is,

a(a-b)(a+b)= -1/1= -1

=>(1-b)(1+b) = -1 [since, a=1]

=> 1- b^2 = -1

=> b^2 = 2

=> b= √2, -√2

Thus, we get, a= 1 & b= √2, -√2.

Hope, it'll help!!

Answer 2

The zeroes of the polynomial $x^3-3x^2+x+1$ are $\dfrac{a}{r},\ a$ and $ar.$

Then the product of the zeroes is given by,

$\longrightarrow\dfrac{a}{r}\cdot a\cdot ar=-\dfrac{1}{1}$

$\longrightarrow a^3=-1$

$\longrightarrow a^3+1=0$

$\longrightarrow a^3+1^3=0$

Since $x^3+y^3=(x+y)(x^2-xy+y^2),$

$\longrightarrow (a+1)(a^2-a+1)=0$

This implies,

$\longrightarrow a=-1$

or,

$\longrightarrow a^2-a+1=0$

$\longrightarrow a=\dfrac{1\pm\sqrt{(-1)^2-4\cdot1\cdot1}}{2\cdot1}$

$\longrightarrow a=\dfrac{1\pm\sqrt{-3}}{2}$

$\longrightarrow a=-\dfrac{-1\pm i\sqrt{3}}{2}\quad\quad\left[i=\sqrt{-1}\right]$

Let $\omega=\dfrac{-1-i\sqrt3}{2}$ and $\omega^2=\dfrac{-1+i\sqrt3}{2}.$ Then,

$\longrightarrow a=-\omega\quad\quad OR\quad\quad a=-\omega^2$

Therefore, possible values for $a$ are,

$\longrightarrow\underline{\underline{a\in\{-1,\ -\omega,\ -\omega^2\}}}$

If $a\in\mathbb{R},$

$\longrightarrow\underline{\underline{a=-1}}$