Question

Form a Quadratic polynomial whose one zero is 3 + V2 and the sum of zeros 6​

Answer 1

EXPLANATION.

Quadratic polynomial whose zeroes = 3 + √2.

Sum of the zeroes of Quadratic polynomial = 6.

As we know that,

One zeroes = 3 + √2.

Other root be their  Conjugate = 3 - √2.

We can also find by this method,

Let one root be = α = 3 + √2.

Sum of zeroes of quadratic equation.

⇒ α + β = 6.

⇒ 3 + √2 + β = 6.

⇒ β = 6 - (3 + √2).

⇒ β = 6 - 3 - √2.

⇒ β = 3 - √2.

As we know that,

Sum of zeroes of quadratic polynomial.

⇒ α + β = -b/a.

⇒ α + β = 6.

Products of zeroes of quadratic polynomial.

⇒ αβ = c/a.

⇒ (3 + √2).(3 - √2).

⇒ (3²) - (√2)².

⇒ 9 - 2.

⇒ 7.

⇒ αβ = 7.

Formula of quadratic equation.

⇒ x² - (α + β)x + αβ.

Put the values in equation, we get.

⇒ x² - (6)x + 7 = 0.

⇒ x² - 6x + 7 = 0.

                                                                                                                           

MORE INFORMATION.

Nature of the factors of the quadratic expression.

(1) = Real and different, if b² - 4ac > 0.

(2) = Rational and different, if b² - 4ac is a perfect square.

(3) = Real and equal, if b² - 4ac = 0.

(4) = If D < 0 Roots are imaginary and unequal or complex  conjugate.

Answer 2

$\rm :\implies\:Let \: \alpha \: and \: \beta \: be \: the \: zeroes \: of \: polynomial.$

Now,

Given that

$\rm :  \implies \: \alpha \: = 3 + \sqrt{2}$

and

$\rm :\implies\: \alpha + \beta = 6$

So,

$\rm :\implies\:3 + \sqrt{2} + \beta = 6$

$\boxed{ \pink{ \rm :\implies\: \: \beta \: = 3 - \sqrt{2} }}$

Now,

$\rm :\implies\: \alpha \beta = \: (3 + \sqrt{2} )(3 - \sqrt{2} )$

$\rm :  \implies \: \alpha \beta \: = {9}^{2} - {( \sqrt{2}) }^{2}$

$\rm :  \implies \: \alpha \beta \: = \: 9 - 2$

$\rm :  \implies \: \alpha \beta \: = \: 7$

So,

Now, we have the values,

$\rm :\implies\: \alpha + \beta = 6$

and

$\rm :  \implies \: \alpha \beta = 7$

Now,

we know,

The quadratic polynomial f(x) is given by

$\rm :  \implies \:f(x) = k( {x}^{2} - ( \alpha + \beta )x + \alpha \beta \: where \: k \: \ne \: 0$

$\boxed{ \pink{\rm :  \implies \:f(x) = k( {x}^{2} - 6x + 7) \: where \: k \: \ne \: 0}}$