Question

Write the quadratic equation whose roots are V5 - 4 and V5+4​

Answer 1

Answer:

x² - 2√5 - 11

Step-by-step explanation:

It is given that √5 - 4 and √5 + 4 are the zeroes of the required polynomial.

Let the two zeroes be α and β of the required polynomial.

∴ α = √5 - 4, β = √5 + 4

_____________________________

Now,

• Sum of zeroes = α + β

→ (√5 - 4) + (√5 + 4)

→ √5 - 4 + √5 + 4

→ √5 + √5

→ 2√5

• Product of zeroes = αβ

→ (√5 - 4)(√5 + 4)

• Identity : (a - b)(a + b) = a² - b²

Here, a = √5, b = 4

→ (√5)² - (4)²

→ 5 - 16

→ - 11

_____________________________

The required polynomial is :

→ p(x) = k [ x² - (α + β)x + αβ ]

• Putting known values.

→ p(x) = k [ x² - (2√5)x + (- 11) ]

→ p(x) = k [x² - 2√5x - 11]

• Putting k = 1.

→ p(x) = x² - 2√5 - 11

___________________________

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https://brainly.in/question/16097680

Answer 2

AnswEr :

$\:\bullet\:\sf\ \alpha = \sqrt{5} - 4$

$\:\bullet\:\sf\ \beta = \sqrt{5} + 4$

$\underline{\dag\:\textsf{According \: to \: given \: in \: question:}}$

★Sum of Roots -

$\normalsize\ : \implies{\boxed{\sf{Sum \: of \: roots = (\alpha + \beta) }}}$

$\normalsize\ : \implies\sf\ \alpha + \beta = (\sqrt{5} - \cancel{4} + \sqrt{5} + \cancel{4}) \\ \\ \normalsize\ : \implies\sf\ \sqrt{5} + \sqrt{5} = 2\sqrt{5}$

$\normalsize\ : \implies{\underline{\boxed{\sf \orange{Sum \: of \: roots(\alpha + \beta) = 2\sqrt{5} }}}}$

$\rule{100}2$

★Product of Roots -

$\normalsize\ : \implies{\boxed{\sf{Product \: of \: roots = \alpha\beta}}} \\ \\ \normalsize\ : \implies\ \alpha\beta = (\sqrt{5} - 4)(\sqrt{5} + 4) \\ \\ \normalsize\ : \implies\sf\ 5 + 4\sqrt{5} - 4\sqrt{5} - 16 \\ \\ \normalsize\ : \implies\sf\ 5 - 16 = -11$

$\normalsize\ : \implies{\underline{\boxed{\sf \orange{Product \: of \: roots(\alpha\beta) = -11}}}}$

$\rule{100}2$

★Now, Quadratic Equation -

$\normalsize\ : \implies{\boxed{\sf{P(x) = k[x^2 + (\alpha + \beta)x - \alpha\beta] }}}$

$\normalsize\ : \implies\sf\ P(x) = k[x^2 - 2\sqrt{5}x + (-11) ]$

$\scriptsize\sf{\: \: \: \: \: \:( \therefore\ \: \red{Here, \: k \: is \: constant \: so, \: k = 1}) }$

$\normalsize\ : \implies\sf\ P(x) = 1[x^2 - 2\sqrt{5}x - 11 ] \\ \\ \normalsize\ : \implies\sf\ P(x) = x^2 - 2\sqrt{5}x - 11$

$\normalsize\ : \implies{\underline{\boxed{\sf \green{P(x) = x^2 - 2\sqrt{5} - 11}}}}$