# If $\alpha + \beta = 5$ and ${\alpha ^3} + {\beta ^3} = 35$ , find the quadratic equation whose roots are $\alpha $ and $\beta .$

Last updated date: 19th Mar 2023

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Answer

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Hint- To find the quadratic equations first, we have to find the product of roots. We will get it with the help of given values. Then we will put the value of sum of roots and product of roots in quadratic formula.

“Complete step-by-step answer:”

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

As we know that

$

{(a + b)^3} = {a^3} + {b^3} + 3ab(a + b) \\

{a^3} + {b^3} = {(a + b)^3} - 3ab(a + b) \\

$

We will write this expression in terms of $\alpha $ and $\beta .$

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

By putting the value of $\alpha + \beta = 5$ and ${\alpha ^3} + {\beta ^3} = 35$ in above equation, we get

$

\Rightarrow 35 = {(5)^3} - 3\alpha \beta (5) \\

\Rightarrow 35 = 125 - 15\alpha \beta \\

\Rightarrow 15\alpha \beta = 90 \\

\Rightarrow \alpha \beta = 6 \\

$

We know that if $\alpha $ and $\beta $ are the roots quadratic equation, then the quadratic equation is

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

On substituting the value of $\alpha + \beta = 5$ and $\alpha \beta = 6$ , we get

\[ \Rightarrow {x^2} - 5x + 6 = 0\]

Hence, the quadratic equation will be \[{x^2} - 5x + 6 = 0\]

Note- To solve questions related to quadratic equations, remember the basic properties of quadratic equations such as sum of roots and product of roots of quadratic equation can be used to form the quadratic equations. Root of quadratic equation all satisfies the quadratic equation and some problems this helps to find the coefficients of quadratic equation.

“Complete step-by-step answer:”

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

As we know that

$

{(a + b)^3} = {a^3} + {b^3} + 3ab(a + b) \\

{a^3} + {b^3} = {(a + b)^3} - 3ab(a + b) \\

$

We will write this expression in terms of $\alpha $ and $\beta .$

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

By putting the value of $\alpha + \beta = 5$ and ${\alpha ^3} + {\beta ^3} = 35$ in above equation, we get

$

\Rightarrow 35 = {(5)^3} - 3\alpha \beta (5) \\

\Rightarrow 35 = 125 - 15\alpha \beta \\

\Rightarrow 15\alpha \beta = 90 \\

\Rightarrow \alpha \beta = 6 \\

$

We know that if $\alpha $ and $\beta $ are the roots quadratic equation, then the quadratic equation is

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

On substituting the value of $\alpha + \beta = 5$ and $\alpha \beta = 6$ , we get

\[ \Rightarrow {x^2} - 5x + 6 = 0\]

Hence, the quadratic equation will be \[{x^2} - 5x + 6 = 0\]

Note- To solve questions related to quadratic equations, remember the basic properties of quadratic equations such as sum of roots and product of roots of quadratic equation can be used to form the quadratic equations. Root of quadratic equation all satisfies the quadratic equation and some problems this helps to find the coefficients of quadratic equation.

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