Question

Form the quadratic polynomial whose zeroes are (7+2root2) and (7-2root2)​

Answer 1

EXPLANATION.

Quadratic polynomial.

Whose zeroes are = (7 + 2√2) and (7 - 2√2).

As we know that,

Sum of the zeroes of the quadratic equation.

⇒ α + β = -b/a.

⇒ (7 + 2√2) + (7 - 2√2).

⇒ 7 + 2√2 + 7 - 2√2 = 14.

⇒ α + β = 14.

Products of the zeroes of the quadratic equation.

⇒ αβ = c/a.

⇒ (7 + 2√2)(7 - 2√2).

As we know that,

Formula of :

⇒ (x + y)(x - y) = x² - y².

Using this formula in equation, we get.

⇒ [(7)² - (2√2)²].

⇒ [49 - 8] = 41.

⇒ αβ = 41.

As we know that,

Formula of quadratic polynomial.

⇒ x² - (α + β)x + αβ.

Put the values in the equation, we get.

⇒ x² - (14)x + (41) = 0.

⇒ x² - 14x + 41 = 0.

                                                                                                                           

MORE INFORMATION.

Conditions for common roots.

Let quadratic equation are : a₁x² + b₁x + c₁ = 0 and a₂x² + b₂x + c₂ = 0.

(1) = If only one root is common.

x = b₁c₂ - b₂c₁/a₁b₂ - a₂b₁.

y = c₁a₂ - c₂a₁/a₁b₂ - a₂b₁.

(2) = If both roots are common.

a₁/a₂ = b₁/b₂ = c₁/c₂.

Answer 2

$\bf \color{navy}Answer:$

$\bf \color{red} \: {x}^{2} - 14x + 41 = 0$

$\bf \color{navy}Step-by-step \: explanation:$

$\bf \color{purple} \alpha = 7 + 2 \sqrt{2}$

$\bf \color{purple} \beta = 7 - 2 \sqrt{2}$

$\bf \color{purple} \alpha + \beta = 14$

$\bf \color{purple} \alpha \: . \: \beta = {(7)}^{2} - (2 \sqrt{2} )^{2}$

$\: \: \: \: \: \: \: \: \: \: \bf \color{purple} = 49 - 8$

$\bf \color{purple} \alpha \: . \: \beta = 41$

$\bf \color{purple} \: Q \: . \: e\: \: \: is$

$\bf \color{purple} {x}^{2} - ( \alpha + \beta )x + \alpha \: . \: \beta = 0$

$\color{navy} = > \bf \color{red} \: {x}^{2} - 14x + 41 = 0$