Question

Find a ^3+ ß^3 , ifa and B are the zeroes of the polynomial
2 x^2 -x + 5​

Answer 1

Answer:-

Given:-

α and β are the roots of 2x² - x + 5.

On comparing the given polynomial with the standard form of a quadratic equation i.e., ax² + bx + c = 0 we get;

a = 2

b = - 1

c = 5.

We know that;

Sum of the roots = - b/a

⟹ α + β = - ( - 1) / 2

⟹ α + β = 1/2 -- equation (1)

And,

product of the roots = c/a

⟹ αβ = 5/2 -- equation (2)

Now,

we have to find:-

⟹ α³ + β³

We know that;

a³ + b³ = (a + b)³ - 3ab(a + b)

Hence;

⟹ α³ + β³ = (α + β)³ - 3αβ(α + β)

Putting the respective values from equations (1) and (2) we get;

⟹ α³ + β³ = (1/2)³ - 3(5/2)(1/2)

⟹ α³ + β³ = (1/8) - (15/4)

⟹ α³ + β³ = (1 - 30)/8

⟹ α³ + β³ = - 29/8

∴ The value of α³ + β³ is - 29/8.

Answer 2

Answer:

Given :-

• α and β are the zeros of the polynomial 2x² - x + 5.

To Find :-

• What is the value of α³ + β³.

Solution :-

Given equation :

$\longmapsto$ $\sf 2{x}^{2} - x + 5 =\: 0$

where,

• a = 2
• b = - 1
• c = 5

Now, we have to find the sum of two roots :

$\mapsto \sf\boxed{\bold{\pink{\alpha + \beta =\: - \dfrac{b}{a}}}}$

Then,

$\implies \sf \alpha + \beta =\: - \dfrac{- 1}{2}$

$\implies \sf\bold{\purple{\alpha + \beta =\: \dfrac{1}{2}}}$

Again, we have to find the product of two roots :

$\mapsto \sf\boxed{\bold{\pink{\alpha\beta =\: \dfrac{c}{a}}}}$

Then,

$\implies \sf\bold{\purple{\alpha\beta =\: \dfrac{5}{2}}}$

Now, we have to find the value of α³ + β³ :

As we know that :

➦ a³ + b³ = (a + b)³ - 3ab(a + b)

According to the question by using the formula we get :

$\leadsto \sf {\alpha}^{3} + {\beta}^{3} =\: {(\alpha + \beta)}^{3} - 3\alpha\beta(\alpha + \beta)$

$\leadsto \sf {\alpha}^{3} + {\beta}^{3} =\: {\bigg(\dfrac{1}{2}\bigg)}^{3} - 3\bigg(\dfrac{5}{2}\bigg)\bigg(\dfrac{1}{2}\bigg)$

$\leadsto \sf {\alpha}^{3} + {\beta}^{3} =\: \dfrac{1}{8} - 3 \times \dfrac{5}{2} \times \dfrac{1}{2}$

$\leadsto \sf {\alpha}^{3} + {\beta}^{3} =\: \dfrac{1}{8} - 3 \times \dfrac{5}{4}$

$\leadsto \sf {\alpha}^{3} + {\beta}^{3} =\: \dfrac{1}{8} - \dfrac{15}{4}$

$\leadsto \sf {\alpha}^{3} + {\beta}^{3} =\: \dfrac{1 - 30}{8}$

$\leadsto \sf\bold{\red{{\alpha}^{3} + {\beta}^{3} =\: \dfrac{- 29}{8}}}$

$\sf\boxed{\bold{\green{\therefore\: The\: value\: of\: {\alpha}^{3} + {\beta}^{3}\: is\: \dfrac{- 29}{8}.}}}$