Question

5) Sum of the roots of the quadratic equation is 5 and sum of their cubes is
35 then find the quadratic equation.​

Answer 1

Answer:

Let the two roots be α and β

α+β=5 , α

3

+β

3

=35

(α+β)

3

=α

3

+β

3

+3αβ(α+β)

(5)

3

=35+3αβ(5)

125=35+15αβ

90=15αβ

∴

15

90

=αβ

∴αβ=6

Quadratic Eq

n

:

x

2

−(sumofroots)x+productofroots=0

x

2

−(α+β)x+αβ=0

x

2

−5x+6=0

Step-by-step explanation:

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Answer 2

Solution:

According to the given condition,

$\alpha + \beta = 5 \\ \\ {\alpha }^{3} + {\beta}^{3} = 35$

But, a³+b³=(a+b)(a²+ab+b²)....{Formula}

So,

$35 = ( \alpha + \beta )( { \alpha }^{2} + \alpha \beta + { \beta }^{2} )$

But,

$( \alpha + \beta ) = 5$

Therefore,

$35 = 5( { \alpha }^{2} + 2 \alpha \beta + { \beta }^{2} ) \\ \\ = > \frac{\cancel{35}}{\cancel5} = 7 =( { \alpha }^{2} + 2 \alpha \beta + { \beta }^{2} )$

Similarly, by splitting (2αβ)

We get,

$7 = ( { \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta - 3 \alpha \beta ) \\ \\ = > 7 = ( { \alpha }^{2} + 2 \alpha \beta + { \beta }^{2} ) - 3 \alpha \beta$

But, α²+2αβ+β²= (α+β)²

...{(a+b)²=a²+2ab+b²}

...{Formula}

Therefore,

$7 = {( \alpha + \beta )}^{2} - 3 \alpha \beta$

But, α+β=5

Therefore,

$7 = {(5)}^{2} - 3 \alpha \beta \\ \\ = > 7 = 25 - 3 \alpha \beta \\ \\ = > 3 \alpha \beta = 25 - 7 \\ \\ = > 3 \alpha \beta = 18 \\ \\ = > \alpha \beta = \frac{\cancel{18}}{\cancel{3}} = 6 \\ \\ = > \alpha \beta = \green{{6}}$

Therefore,

The equation will be..

${x}^{2} - ( \alpha + \beta )x + (\alpha \beta ) = 0$

...{By Formula}

$\green{{x}^{2} - 5x + 6 = 0}$

Therefore, following is the required equation.