Question

if alpha and beta are the two zeroes of x²-4x-1 ,find alpha-beta , alpha²+beta² , alpha³+beta³, 1/alpha + 1/beta​

Answer 1

Answer:

18, 76, - 4

Step-by-step explanation:

This is a quadratic polynomial in form of ax² + bx + c. In such case, if α and β are roots,

Sum of roots = -b/a → α + β = - b/a

Product of roots = c/a → αβ = c/a

In the given question,

α + β = - (-4)/1 = 4

αβ = (-1)/1 = - 1

Therefore,

(i):

α² + β² = (α + β)² - 2αβ = (4)² - 2(-1) = 18

(ii):

α³+β³ = (α + β)(α² + β² - αβ) = (4)(18-(-1))=76

(iii):

1/α + 1/β = (α + β)/αβ = (4)/(-1) = - 4

Answer 2

Answer:

Given :-

$\leadsto$ The two zeros of x² - 4x - 1.

To Find :-

$\leadsto$ What is the value of :

• α² + β²
• α³ + β³
• 1/α + 1/β

Solution :-

Given equation :

$\mapsto \sf\bold{\green{{x}^{2} - 4x - 1}}$

where,

• a = 1
• b = - 4
• c = - 1

Now, as we know that :

$\clubsuit$ Sum of roots :

$\longmapsto \sf\boxed{\bold{\pink{Sum\: of\: roots\: (\alpha + \beta) =\: \dfrac{- b}{a}}}}$

Given :

• b = - 4
• a = 1

According to the question by using the formula we get

$\implies \sf \alpha + \beta =\: \dfrac{-(- 4)}{1}$

$\implies \sf \alpha + \beta =\: \dfrac{4}{1}$

$\implies \sf\bold{\purple{\alpha + \beta =\: 4}}$

Again, we know that,

$\clubsuit$ Product of roots :

$\longmapsto \sf\boxed{\bold{\pink{Product\: of\: roots\: (\alpha\beta) =\: \dfrac{c}{a}}}}$

Given :

• c = - 1
• a = 1

According to the question by using the formula we get,

$\implies \sf \alpha\beta =\: \dfrac{- 1}{1}$

$\implies \sf\bold{\purple{\alpha\beta =\: - 1}}$

$\rule{150}{2}$

Now,

$\rightarrow \sf\bold{\blue{{\alpha}^{2} + {\beta}^{2}}}$

As we know that :

$\longmapsto \sf\boxed{\bold{\pink{{a}^{2} + {b}^{2} =\: {(a + b)}^{2} - 2ab}}}$

Then, by using the formula we get,

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

$\implies \sf {\alpha}^{2} + {\beta}^{2} =\: {(4)}^{2} - 2(- 1)$

$\implies \sf {\alpha}^{2} + {\beta}^{2} =\: 16 + 2$

$\implies\sf\bold{\red{{\alpha}^{2} + {\beta}^{2} =\: 18}}$

$\rule{150}{2}$

$\rightarrow \sf\bold{\blue{{\alpha}^{3} + {\beta}^{3}}}$

As we know that,

$\longmapsto \sf\boxed{\bold{\pink{{a}^{3} + {b}^{3} =\: (a + b)({a}^{2} + {b}^{2} - ab)}}}$

Then, by using the formula we get,

As we already get the value of α² + β² = 18, so

$\implies \sf {\alpha}^{3} + {\beta}^{3} =\: (4)(18 - \{- 1\})$

$\implies \sf {\alpha}^{3} + {\beta}^{3} =\: (4)(18 + 1)$

$\implies \sf {\alpha}^{3} + {\beta}^{3} =\: 4 \times 19$

$\implies \sf\bold{\red{{\alpha}^{3} + {\beta}^{3} =\: 76}}$

$\rule{150}{2}$

$\rightarrow \sf\bold{\blue{\dfrac{1}{\alpha} + \dfrac{1}{\beta}}}$

$\implies \sf \dfrac{1}{\alpha} + \dfrac{1}{\beta} =\: \dfrac{\beta + \alpha}{\alpha\beta}$

$\implies \sf \dfrac{1}{\alpha} + \dfrac{1}{\beta} =\: \dfrac{\alpha + \beta}{\alpha\beta}$

$\implies \sf \dfrac{1}{\alpha} + \dfrac{1}{\beta} =\: \dfrac{4}{- 1}$

$\implies \sf\bold{\red{\dfrac{1}{\alpha} + \dfrac{1}{\beta} =\: - 4}}$

$\rule{150}{2}$