Question

if the zero of polynomial x square minus 3 X square + X + 1 are (a +b), a, (a+b) find the value of a and b​

Answer 1

hi

buddy ✌️✌️✌️✌️

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u r question is incorrect it will be x cube minus 3 x square

❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️

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@toshika

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Answer 2

GIVEN :

The zeros of polynomial $x^3-3x^2+x+1$  are (a -b), a, (a+b)

TO FIND :

The values of a and b

SOLUTION :

Given polynomial is $x^3 - 3 x^2 + x + 1$

Equate the polynomial to zero to find the zeros

By using the Synthetic Division Method

1_|  1     -3      1       1

     0      1     -2      -1

   ______________

     1     -2     -1        0

 

∴ x-1 is a factor

x-1=0

∴ x=1 is a zero

Now we have the quadratic equation $x^2-2x-1=0$

The formula for the quadratic equation for x is :

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Comparing the coefficients in quadratic equation $x^2-2x-1=0$ with the general quadratic equation we get a=1 , b=-2 and c=-1

Now substitute the values in the formula we have

$x=\frac{-(-2)\pm \sqrt{(-2)^2-4(1)(-1)}}{2(1)}$

$=\frac{2\pm \sqrt{4+4}}{2}$

$=\frac{2\pm \sqrt{8}}{2}$

$=\frac{2\pm 2\sqrt{2}}{2}$

$x=1\pm \sqrt{2}$

∴ $x=1+\sqrt{2}$ and $x=1-\sqrt{2}$ are the zeros

∴ the zeros are 1 , $1+\sqrt{2}$ and $1-\sqrt{2}$

Comparing with the given zeros a, a+b and a-b with the zeros

1 , $1+\sqrt{2}$ and $1-\sqrt{2}$ we get

Hence we get a=1 and $b=\sqrt{2}$

∴ the value of a is 1 and b is $\sqrt{2}$