Question:

If α,β,γ \alpha ,\beta ,\gamma are the roots of the equation x37x+7=0, {{x}^{3}}-7x+7=0, then 1α4+1β4+1γ4 \frac{1}{{{\alpha }^{4}}}+\frac{1}{{{\beta }^{4}}}+\frac{1}{{{\gamma }^{4}}} is

Updated On: Jun 23, 2024
  • 73 \frac{7}{3}
  • 37 \frac{3}{7}
  • 47 \frac{4}{7}
  • 74 \frac{7}{4}
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The Correct Option is B

Solution and Explanation

Given α,β,γ \alpha ,\beta ,\gamma
are the roots of the equation
x37x+7=0 {{x}^{3}}-7x+7=0 .
\therefore Σα=0,  Σαβ=7,  αβγ=7 \Sigma \alpha =0,\,\,\Sigma \alpha \beta =-7,\,\,\alpha \beta \gamma =-7
Now, 1α4+1β4+1γ4=α4β4+β4γ4+γ4α4α4β4γ4 \frac{1}{{{\alpha }^{4}}}+\frac{1}{{{\beta }^{4}}}+\frac{1}{{{\gamma }^{4}}}=\frac{{{\alpha }^{4}}{{\beta }^{4}}+{{\beta }^{4}}{{\gamma }^{4}}+{{\gamma }^{4}}{{\alpha }^{4}}}{{{\alpha }^{4}}{{\beta }^{4}}{{\gamma }^{4}}}
=Σα4β4α4β4γ4 =\frac{\Sigma {{\alpha }^{4}}{{\beta }^{4}}}{{{\alpha }^{4}}{{\beta }^{4}}{{\gamma }^{4}}} ..(i)
ΣαβΣαβΣαβΣαβ=(Σαβ)2.(Σαβ)2 \Sigma \alpha \beta \,\Sigma \alpha \beta \,\Sigma \alpha \beta \,\Sigma \alpha \beta ={{(\Sigma \alpha \beta )}^{2}}.{{(\Sigma \alpha \beta )}^{2}}
\Rightarrow (7)4=(α2β2+β2γ2+γ2α2+2α2βγ {{(-7)}^{4}}=({{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}+{{\gamma }^{2}}{{\alpha }^{2}}+2{{\alpha }^{2}}\beta \gamma
+2αβ2γ+2αβγ2) +2\alpha {{\beta }^{2}}\gamma +2\alpha \beta {{\gamma }^{2}})
(α2β2+β2γ2+γ2α2+2α2β2+2αβ2γ+2αβγ2) ({{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}+{{\gamma }^{2}}{{\alpha }^{2}}+2{{\alpha }^{2}}{{\beta }^{2}}+2\alpha {{\beta }^{2}}\gamma +2\alpha \beta {{\gamma }^{2}})
=[α2β2+β2γ2+γ2α2+2αβγ(α+β+γ)] =[{{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}+{{\gamma }^{2}}{{\alpha }^{2}}+2\alpha \beta \gamma (\alpha +\beta +\gamma )]
[α2β2+β2γ2+γ2α2+2αβγ(α+β+γ)] [{{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}+{{\gamma }^{2}}{{\alpha }^{2}}+2\alpha \beta \gamma (\alpha +\beta +\gamma )]
=(α2β2+β2γ2+γ2α2) =({{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}+{{\gamma }^{2}}{{\alpha }^{2}})
(α2β2+β2γ2+γ2α2) ({{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}+{{\gamma }^{2}}{{\alpha }^{2}})
(  Σα=α+β+γ=0) (\because \,\,\Sigma \alpha =\alpha +\beta +\gamma =0)
=α2β2+β2γ4+γ4α4+2α4β2γ2 ={{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{4}}+{{\gamma }^{4}}{{\alpha }^{4}}+2{{\alpha }^{4}}{{\beta }^{2}}{{\gamma }^{2}}
+2α2β4γ2+2α2β2γ4 +2{{\alpha }^{2}}{{\beta }^{4}}{{\gamma }^{2}}+2{{\alpha }^{2}}{{\beta }^{2}}{{\gamma }^{4}}
=Σα2β4+2α2β2γ2(α2+β2+γ2) =\Sigma {{\alpha }^{2}}{{\beta }^{4}}+2{{\alpha }^{2}}{{\beta }^{2}}{{\gamma }^{2}}({{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}})
=Σα4β4+2α2β2γ2[(Σα)22Σαβ] =\Sigma {{\alpha }^{4}}{{\beta }^{4}}+2{{\alpha }^{2}}{{\beta }^{2}}{{\gamma }^{2}}[{{(\Sigma \alpha )}^{2}}-2\Sigma \alpha \beta ]
=Σα4β4+2α2β2γ2[02×(7)] =\Sigma {{\alpha }^{4}}{{\beta }^{4}}+2{{\alpha }^{2}}{{\beta }^{2}}{{\gamma }^{2}}[0-2\times (-7)]
\Rightarrow (7)4=Σα4β4+2(7)2(2×7) {{(-7)}^{4}}=\Sigma {{\alpha }^{4}}{{\beta }^{4}}+2{{(-7)}^{2}}(2\times 7)
\Rightarrow Σα4β4=(7)4+4(7)3 \Sigma {{\alpha }^{4}}{{\beta }^{4}}={{(-7)}^{4}}+4{{(-7)}^{3}}
\Rightarrow Σα2β4=(7)3(7+4)=3(7)3 \Sigma {{\alpha }^{2}}{{\beta }^{4}}={{(-7)}^{3}}(-7+4)=-3{{(-7)}^{3}}
On putting this value in E (i), we get 1α4+1β4+1γ4=3(7)3(7)4=37=37 \frac{1}{{{\alpha }^{4}}}+\frac{1}{{{\beta }^{4}}}+\frac{1}{{{\gamma }^{4}}}=\frac{-3{{(-7)}^{3}}}{{{(-7)}^{4}}}=\frac{-3}{-7}=\frac{3}{7} `
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Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.

Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.