Question

Find the quadratic polynomials whose zeroes are
given below
3-V2, 3+2​

Answer 1

Question :

Find the quadratic polynomials whose zeroes are 3 - √2 and 3 + √2.

Given :

• First zero of the Quadratic polynomial = 3 - √2.

• Second zero of the Quadratic polynomial = 3 + √2.

Solution :

Let the First zero of Quadratic polynomial be α and the second zero be β.

We know the formula for Quadratic polynomial,i.e,

$\boxed{\bf{x^{2} + (\alpha + \beta)x + \alpha\beta}}$

Now by using the above formula and substituting the values in it, we get :

Here ,

• α = 3 - √2
• β = 3 + √2

$\boxed{\bf{x^{2} + (\alpha + \beta)x + \alpha\beta}}$

$:\implies \bf{x^{2} + (\alpha + \beta)x + \alpha\beta}$

$:\implies \bf{x^{2} + (3 + \sqrt{2} + 3 - \sqrt{2})x + (3 + \sqrt{2})(3 - \sqrt{2})}$

$:\implies \bf{x^{2} + (3 + \not{\sqrt{2}} + 3 - \not{\sqrt{2}})x + (3 + \sqrt{2})(3 - \sqrt{2})}$

$:\implies \bf{x^{2} + (3 + 3)x + (3 + \sqrt{2})(3 - \sqrt{2})}$

$:\implies \bf{x^{2} + 6x + (3 + \sqrt{2})(3 - \sqrt{2})}$

$:\implies \bf{x^{2} + 6x + 3(3 - \sqrt{2}) + \sqrt{2}(3 - \sqrt{2})}$

$:\implies \bf{x^{2} + 6x + 9 - 3\sqrt{2} + 3\sqrt{2} - 2}$

$:\implies \bf{x^{2} + 6x + 9 - \not{3\sqrt{2}} + \not{3\sqrt{2}} - 2}$

$:\implies \bf{x^{2} + 6x + 9 - 2}$

$:\implies \bf{x^{2} + 6x + 7}$

$\boxed{\therefore \bf{x^{2} + 6x + 7}}$

Hence the quadratic equation formed is x² + 6x + 7.

Answer 2

Question

Find the quadriatic whose zeroes are given below 3 - √2 , 3 + √2

Given

• first zero of quadriatic polynomial 3 - √2
• second zero of quadriatic polynomial 3 + √2

Solution

Given zeros of polynomial are 3 + √2

$\sf \: let \: a \: and \: \beta \: be \: zeros \: of \: polynomial \:$

$\\ \sf \: a + \beta = 3 + \sqrt{2} + 3 - \sqrt{2} = 6..........(1)$

$\\ \sf \: a \beta (3 + \sqrt{2} )(3 - \sqrt{2} )$

$\\ \sf \: a \beta = 9 - 3 \sqrt{2} + 3 \sqrt{2} - 2 = 7..........(2)$

then quadriatic equation is

$\\ \sf {x}^{2} - (a + \beta )x + a \beta = 0$

from (1) and (2)

$\\ \sf \: hence \: the \: quadriatic \: equation \: we \: get \: {x }^{2} - 6x + 7 = 0$