Question

If alpha and beta are the zeroes of
x^2- 3x-2. Find a quadratic polynomial whose zeroes are: the pic is given below

Answer 1

$\huge{\boxed{\mathbb{Answer\::-}}}$

As it's given that $\alpha$ and $\beta$ are the roots of the quadratic equation

$x^2 -3x -2$

♦ By using Quadratic formula for finding roots .

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

$= \dfrac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-2)}}{2(1)}$

$= \dfrac{3 \pm \sqrt{9 + 8}}{2}$

$= \dfrac{3 \pm \sqrt{17}}{2}$

♦ So let

$\alpha = \dfrac{3 + \sqrt{17}}{2}$

and

$\beta = \dfrac{3 - \sqrt{17}}{2}$

♦ Now we will first find out the value of

>> (i)

$\dfrac{1}{2\alpha+\beta} \\\\ = \dfrac{1}{2\dfrac{3+\sqrt{17}}{2} + \dfrac{3-\sqrt{17}}{2}}\\\\=\dfrac{1}{\dfrac{6+2\sqrt{17}}{2} + \dfrac{3-\sqrt{17}}{2}}\\\\ = \dfrac{1}{9+\sqrt{17}}{2} \\\\ = \dfrac{2}{9+\sqrt{17}}$

• On rationalizing denominator

$\dfrac{2}{9+\sqrt{17}} \times \dfrac{9-\sqrt{17}}{9-\sqrt{17}}$

$= \dfrac{18 - 2\sqrt{17}}{9^2 - \sqrt{17}^2 }$

$= \dfrac{18-2\sqrt{17}}{81-17} \\\\ = \dfrac{18-2\sqrt{17}}{64} \\\\ = \dfrac{9-\sqrt{17}}{32}$

>> (ii)

$= \dfrac{1}{2\beta + \alpha}$

$= \dfrac{1}{2\dfrac{3-\sqrt{17}}{2} + \dfrac{3+\sqrt{17}}{2}}$

$= \dfrac{1}{\dfrac{6-2\sqrt{17}}{2} + \dfrac{3+\sqrt{17}}{2}}$

$= \dfrac{1}{\dfrac{9-\sqrt{17}}{2}}$

$= \dfrac{2}{9-\sqrt{17}}$

• On rationalizing denominator

$\dfrac{2}{9-\sqrt{17}} \times \dfrac{9+\sqrt{17}}{9+\sqrt{17}}$

$= \dfrac{18 + 2\sqrt{17}}{9^2 - \sqrt{17}^2 }$

$= \dfrac{18+2\sqrt{17}}{81-17} \\\\ = \dfrac{18+2\sqrt{17}}{64} \\\\ = \dfrac{9+\sqrt{17}}{32}$

♦ So by using sum and Product of roots

$x^2 - (\alpha + \beta)x + (\alpha\beta)$

$= x^2 - \dfrac{18}{32}x + \dfrac{64}{1024}$

$= x^2 - \dfrac{9}{16}x + \dfrac{1}{16}$

Answer 2

Solution:-

Given Eq:- x² - 3x - 2

Roots of the Given Equation is Alpha & Beta.

Sum of Roots,

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

Product of Roots,

Alpha × Beta = c/a = -2

Now,

The zeroes of Required Quadratic Polynomial are:-

1/(2a+b) & 1/(a+2b).

Now,

Sum of Roots,

1/(2a+b) + 1/(a+2b)

=> (2b + a+ 2a + b)/(2a+b)(a+2b)

=> 3(a+b)/ [ 2a²+ 4ab + ab + 2b²]

Putting the values (a+b =3) & (ab = -2).

=> 3(3)/ [ 5(-2) + 2[ (a+b)² - 2ab]

=> 9 / [ -10 + 2(9 + 4)]

=> 9/ 16.

Now, Product of Zeroes,

1/(2a+b) × 1/(a+2b)

=> 1/(2a+b)(a+2b)

=> 1/(4ab + 2a² + 2b² + ab)

=> 1/[ 5(-2) + 2(a+b)² - 2ab]

=> 1/[ -10 + 2(9+4)]

=> 1/[ -10 + 26]

=> 1/16.

Now,

By Quadratic Polynomial Equation:-

x² - (sum of Roots)x + ( product of roots)

=> x² - 9/16x + 1/16

Hence,

x² - 9/16x + 1/16 is the required Quadratic Equation.