
If $\alpha $ and $\beta $ are the roots of the quadratic equation ${{x}^{2}}-4x+1=0$, then find the values of \[{{\alpha }^{3}}+{{\beta }^{3}}\]
Answer
162.9k+ views
Hint: In this question we have given the quadratic equation and we have to find the value of \[{{\alpha }^{3}}+{{\beta }^{3}}\]. For this first we have to find the sum of roots and the product of roots .then we put these values in the formula of cube root and get the desirable answer.
Formula Used:
Sum of roots = $\dfrac{-b}{a}$
Product of roots = $\dfrac{c}{a}$
\[{{a}^{3}}+{{b}^{3}}={{(a+b)}^{3}}-3ab(a+b)\]
Complete step by step Solution:
Given quadratic equation is ${{x}^{2}}-4x+1=0$…………………………….. (1)
By comparing the equation (1) with the quadratic equation $a{{x}^{2}}+bx+c=0$
We are able to get a = 1, b = -4, c = +1
Let $\alpha $and $\beta $are the roots of the given quadratic equation.
Then we know general quadratic equation with the roots $\alpha $and $\beta $be
${{x}^{2}}-(\alpha +\beta )x+\alpha \beta =0$………………………………….. (2)
We know in quadratic equation
Sum of roots = $\dfrac{-b}{a}$ = $\dfrac{-(-4)}{1}$ = 4
And the product of roots = $\dfrac{c}{a}$= \[\dfrac{+1}{1}\]= 1
Now we compare the equation (1) and (2), we get
\[\alpha +\beta \]= $\dfrac{-b}{a}$= 4
And \[\alpha \beta \]= $\dfrac{c}{a}$= 1
Now we know the formula of \[{{a}^{3}}+{{b}^{3}}={{(a+b)}^{3}}-3ab(a+b)\]
By putting a = $\alpha $ and b = $\beta $, we get
\[{{\alpha }^{3}}+{{\beta }^{3}}={{(\alpha +\beta )}^{3}}-3\alpha \beta (\alpha +\beta )\]
By putting the values, we get
\[{{\alpha }^{3}}+{{\beta }^{3}}={{(4)}^{3}}-3(1)(4)\]
= 52
Hence, the value of \[{{\alpha }^{3}}+{{\beta }^{3}}=52\]
Note: Whenever we solve these types of questions where roots are given, we use the identity of the product of roots which is if x and y are the roots of any quadratic equation the value of xy will be equal to $\dfrac{ constant\, term}{coefficient\, of\, x^2}$ and sum of the roots that is x + y is equal to $\dfrac{ -coefficient\, of \,x}{coefficient\, of\, x^2}$ and by solving it we get the desired answer.
Formula Used:
Sum of roots = $\dfrac{-b}{a}$
Product of roots = $\dfrac{c}{a}$
\[{{a}^{3}}+{{b}^{3}}={{(a+b)}^{3}}-3ab(a+b)\]
Complete step by step Solution:
Given quadratic equation is ${{x}^{2}}-4x+1=0$…………………………….. (1)
By comparing the equation (1) with the quadratic equation $a{{x}^{2}}+bx+c=0$
We are able to get a = 1, b = -4, c = +1
Let $\alpha $and $\beta $are the roots of the given quadratic equation.
Then we know general quadratic equation with the roots $\alpha $and $\beta $be
${{x}^{2}}-(\alpha +\beta )x+\alpha \beta =0$………………………………….. (2)
We know in quadratic equation
Sum of roots = $\dfrac{-b}{a}$ = $\dfrac{-(-4)}{1}$ = 4
And the product of roots = $\dfrac{c}{a}$= \[\dfrac{+1}{1}\]= 1
Now we compare the equation (1) and (2), we get
\[\alpha +\beta \]= $\dfrac{-b}{a}$= 4
And \[\alpha \beta \]= $\dfrac{c}{a}$= 1
Now we know the formula of \[{{a}^{3}}+{{b}^{3}}={{(a+b)}^{3}}-3ab(a+b)\]
By putting a = $\alpha $ and b = $\beta $, we get
\[{{\alpha }^{3}}+{{\beta }^{3}}={{(\alpha +\beta )}^{3}}-3\alpha \beta (\alpha +\beta )\]
By putting the values, we get
\[{{\alpha }^{3}}+{{\beta }^{3}}={{(4)}^{3}}-3(1)(4)\]
= 52
Hence, the value of \[{{\alpha }^{3}}+{{\beta }^{3}}=52\]
Note: Whenever we solve these types of questions where roots are given, we use the identity of the product of roots which is if x and y are the roots of any quadratic equation the value of xy will be equal to $\dfrac{ constant\, term}{coefficient\, of\, x^2}$ and sum of the roots that is x + y is equal to $\dfrac{ -coefficient\, of \,x}{coefficient\, of\, x^2}$ and by solving it we get the desired answer.
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