Question

if a,b are the zeros of the polynomial 3x^2-4x+1 then find the value of polynomial with zeros are a^2/b and b^2/a​

Answer 1

Answer:

a=1/3    b=1    a^2/b=1/9    b^2/a=3

Step-by-step explanation:

3x^2-4x+1=0

3x^2-3x-x+1=0

3x(x-1)-1(x-1)=0

(3x-1)(x-1)=0

x=1/3 and 1   a=1/3 and b=1

a^2/b=(1/3)^2/1

          =1/9

b^2/a=(1)^2/(1/3)

        =1/(1/3)

        =3

Answer 2

Answer :

• Hence the quadratic equation formed is 3x² - 10x + 1.

Given :

• Given quadratic equation, 3x² - 4x + 1.

To find :

• The quadratic equation whose roots are a²/b and b²/a.

Knowlwdge required :

• Quadratic equation formula :

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

• Formula to form a quadratic equation :

$\boxed{\therefore \sf{x^{2} - (\alpha + \beta)x + \alpha\beta = 0}}$

Solution :

By using the quadratic equation Formula and substituting the values in it, we get :

$:\implies \sf{x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}}$

$:\implies \sf{x = \dfrac{-(-4) \pm \sqrt{(-4)^{2} - 4 \times 3 \times 1}}{2 \times 3}}$

$:\implies \sf{x = \dfrac{4 \pm \sqrt{16 - 12}}{2 \times 3}}$

$:\implies \sf{x = \dfrac{4 \pm \sqrt{4}}{2 \times 3}}$

$:\implies \sf{x = \dfrac{4 \pm 2}{2 \times 3}}$

$:\implies \sf{x = \dfrac{4 \pm 2}{6}}$

$:\implies \sf{x = \dfrac{4 + 2}{6}\:;\:x = \dfrac{4 - 2}{6}}$

$:\implies \sf{x = \dfrac{6}{6}\:;\:x = \dfrac{2}{6}}$

$:\implies \sf{x = \dfrac{\not{6}}{\not{6}}\:;\:x = \dfrac{\not{2}}{\not{6}}}$

$:\implies \sf{x = 1\:;\:x = \dfrac{1}{3}}$

$\boxed{\therefore \sf{x = 1;\dfrac{1}{3}}}$

Thus, the value of x is 1 and ⅓.

From the above equation, we get :

• The first root of the equation, a = 1
• The second root of the equation, b = ⅓

But we are have to find the quadratic equation whose roots are a²/b and b²/a, So first let's solve them.

$:\implies \sf{\alpha = \dfrac{a^{2}}{b} \quad | \quad \beta = \dfrac{b^{2}}{a}}$

Now by substituting the values of a and b in the equation, we get : [Here, a = 1 and b = ⅓]

$:\implies \sf{\alpha = \dfrac{1^{2}}{\dfrac{1}{3}} \quad | \quad \beta = \dfrac{\dfrac{1}{3^{2}}}{1}}$

$:\implies \sf{\alpha = \dfrac{1}{\dfrac{1}{3}} \quad | \quad \beta = \dfrac{\dfrac{1}{9}}{1}}$

$:\implies \sf{\alpha = 3 \quad | \quad \beta = \dfrac{1}{9}}$

$\boxed{\therefore \sf{\alpha = 3 \quad | \quad \beta = \dfrac{1}{9}}}$

Hence the roots of the equation are 3 and ⅑.

Now by using the equation for forming a quadratic equation and substituting the values in it, we get :

$:\implies \sf{x^{2} - (\alpha + \beta)x + \alpha\beta = 0}$

$:\implies \sf{x^{2} - \bigg(3 + \dfrac{1}{9}\bigg)x + 3 \times \dfrac{1}{9} = 0}$

$:\implies \sf{x^{2} - \bigg(\dfrac{27 + 1}{9}\bigg)x + \not{3} \times \dfrac{1}{\not{9}} = 0}$

$:\implies \sf{x^{2} - \dfrac{30x}{9} + \dfrac{1}{3} = 0}$

$:\implies \sf{x^{2} - \dfrac{10x}{3} + \dfrac{1}{3} = 0}$

By multiplying the whole equation by 3, we get :

$:\implies \sf{\bigg(x^{2} - \dfrac{10x}{3} + \dfrac{1}{3}\bigg) \times 3 = 0 \times 3}$

$:\implies \sf{x^{2} \times 3 - \dfrac{10x}{3} \times 3 + \dfrac{1}{3} \times 3 = 0}$

$:\implies \sf{x^{2} \times 3 - \dfrac{10x}{\not{3}} \times \not{3} + \dfrac{1}{\not{3}} \times \not{3} = 0}$

$:\implies \sf{3x^{2} - 10x+ 1 = 0}$

$\boxed{\therefore \sf{3x^{2} - 10x+ 1 = 0}}$

Hence the quadratic equation formed is 3x² - 10x + 1.