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If $\alpha +\beta =-2$ and ${{\alpha }^{3}}+{{\beta }^{3}}=-56$, then the quadratic expression whose roots are $\alpha $ and $\beta $ is :
[a] ${{x}^{2}}+2x-16$
[b] ${{x}^{2}}+2x+15$
[c] ${{x}^{2}}+2x-12$
[d] ${{x}^{2}}+2x-8$
Answer
![Verified]()
Verified
Verified
Hint: Use the formula ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$ and ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ to find the value of $\alpha \beta $ and the use the property that the quadratic expression with roots a and b is ${{x}^{2}}-\left( a+b \right)x+ab$.
Complete step-by-step answer:
We have ${{\alpha }^{3}}+{{\beta }^{3}}=-56$ and $\alpha +\beta =-2$
Using ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$, we get
$\left( \alpha +\beta \right)\left( {{\alpha }^{2}}-\alpha \beta +{{\beta }^{2}} \right)=-56$
Substituting the value of $\left( \alpha +\beta \right)$, we get
$-2\left( {{\alpha }^{2}}-\alpha \beta +{{\beta }^{2}} \right)=-56$
Using ${{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab$ , we get
$-2\left( {{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta -\alpha \beta \right)=-56$
Substituting the value of $\left( \alpha +\beta \right)$, we get
$\begin{align}
& -2\left( {{\left( -2 \right)}^{2}}-2\alpha \beta -\alpha \beta \right)=-56 \\
& \Rightarrow -2\left( 4-3\alpha \beta \right)=-56 \\
\end{align}$
Dividing both sides by -2, we get
$\begin{align}
& \dfrac{-2\left( 4-3\alpha \beta \right)}{-2}=\dfrac{-56}{-2} \\
& \Rightarrow 4-3\alpha \beta =28 \\
\end{align}$
Subtracting 4 on both sides, we get
$\begin{align}
& 4-3\alpha \beta -4=28-4 \\
& \Rightarrow -3\alpha \beta =24 \\
\end{align}$
Dividing both sides by -3, we get
$\begin{align}
& \dfrac{-3\alpha \beta }{-3}=\dfrac{24}{-3} \\
& \Rightarrow \alpha \beta =-8 \\
\end{align}$
Using the property “The quadratic expression with roots a and b is ${{x}^{2}}-\left( a+b \right)x+ab$.”
We have the quadratic expression with roots $\alpha $ and $\beta $ is
$\begin{align}
& {{x}^{2}}-\left( -2 \right)x+\left( -8 \right) \\
& ={{x}^{2}}+2x-8 \\
\end{align}$
Hence option [d] is correct.
Note: [1] We can solve the above question using newton method also.
Let ${{P}_{n}}={{\alpha }^{n}}+{{\beta }^{n}}$ where $\alpha $ and $\beta $ are the roots of the equation ${{x}^{2}}+ax+b$ then we have
${{P}_{1}}=-a$,${{P}_{2}}+a{{P}_{1}}+2b=0$ and ${{P}_{n}}+a{{P}_{n-1}}+b{{P}_{n-2}}=0\forall n\ge 3$
Using we get
${{P}_{1}}=-a$
But ${{P}_{1}}=\alpha +\beta =-2$ we have -a = -2
Hence a = 2.
${{P}_{2}}+a{{P}_{1}}+2b=0$
Substituting the value of ${{P}_{1}}$ and “a” we get
$\begin{align}
& {{P}_{2}}+\left( 2 \right)\left( -2 \right)+2b=0 \\
& \Rightarrow {{P}_{2}}-4+2b=0 \\
\end{align}$
Transposing -4+2b to RHS we get
${{P}_{2}}=4-2b$
Put n = 3 in the recurrence ${{P}_{n}}+a{{P}_{n-1}}+b{{P}_{n-2}}=0$, we get
${{P}_{3}}+\left( 2 \right){{P}_{2}}+b{{P}_{1}}=0$
But ${{P}_{3}}={{\alpha }^{3}}+{{\beta }^{3}}=-56$
Substituting the value of ${{P}_{1}},{{P}_{2}}$ and ${{P}_{3}}$ we get
$\begin{align}
& -56+\left( 2 \right)\left( 4-2b \right)+b\left( -2 \right)=0 \\
& \Rightarrow -56+8-4b-2b=0 \\
& \Rightarrow -48-6b=0 \\
\end{align}$
Adding 6b on both sides, we get
$\begin{align}
& -48-6b+6b=0+6b \\
& \Rightarrow 6b=-48 \\
\end{align}$
Dividing both sides by 6, we get
$\begin{align}
& \dfrac{6b}{6}=\dfrac{-48}{6}=-8 \\
& \Rightarrow b=-8 \\
\end{align}$
Hence b = -8 and a = 2
Hence the quadratic expression is ${{x}^{2}}+ax+b={{x}^{2}}+2x-8$
[2] The derivation of the Newton's method is a direct result of the observation that if $\alpha $ is a root of quadratic expression ${{x}^{2}}+ax+b$ then ${{\alpha }^{n}}+a{{\alpha }^{n-1}}+b{{\alpha }^{n-2}}=0\forall n\ge 3$. Write a similar expression for $\beta $ and add the two equations to get the above recursive relation.
Complete step-by-step answer:
We have ${{\alpha }^{3}}+{{\beta }^{3}}=-56$ and $\alpha +\beta =-2$
Using ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$, we get
$\left( \alpha +\beta \right)\left( {{\alpha }^{2}}-\alpha \beta +{{\beta }^{2}} \right)=-56$
Substituting the value of $\left( \alpha +\beta \right)$, we get
$-2\left( {{\alpha }^{2}}-\alpha \beta +{{\beta }^{2}} \right)=-56$
Using ${{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab$ , we get
$-2\left( {{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta -\alpha \beta \right)=-56$
Substituting the value of $\left( \alpha +\beta \right)$, we get
$\begin{align}
& -2\left( {{\left( -2 \right)}^{2}}-2\alpha \beta -\alpha \beta \right)=-56 \\
& \Rightarrow -2\left( 4-3\alpha \beta \right)=-56 \\
\end{align}$
Dividing both sides by -2, we get
$\begin{align}
& \dfrac{-2\left( 4-3\alpha \beta \right)}{-2}=\dfrac{-56}{-2} \\
& \Rightarrow 4-3\alpha \beta =28 \\
\end{align}$
Subtracting 4 on both sides, we get
$\begin{align}
& 4-3\alpha \beta -4=28-4 \\
& \Rightarrow -3\alpha \beta =24 \\
\end{align}$
Dividing both sides by -3, we get
$\begin{align}
& \dfrac{-3\alpha \beta }{-3}=\dfrac{24}{-3} \\
& \Rightarrow \alpha \beta =-8 \\
\end{align}$
Using the property “The quadratic expression with roots a and b is ${{x}^{2}}-\left( a+b \right)x+ab$.”
We have the quadratic expression with roots $\alpha $ and $\beta $ is
$\begin{align}
& {{x}^{2}}-\left( -2 \right)x+\left( -8 \right) \\
& ={{x}^{2}}+2x-8 \\
\end{align}$
Hence option [d] is correct.
Note: [1] We can solve the above question using newton method also.
Let ${{P}_{n}}={{\alpha }^{n}}+{{\beta }^{n}}$ where $\alpha $ and $\beta $ are the roots of the equation ${{x}^{2}}+ax+b$ then we have
${{P}_{1}}=-a$,${{P}_{2}}+a{{P}_{1}}+2b=0$ and ${{P}_{n}}+a{{P}_{n-1}}+b{{P}_{n-2}}=0\forall n\ge 3$
Using we get
${{P}_{1}}=-a$
But ${{P}_{1}}=\alpha +\beta =-2$ we have -a = -2
Hence a = 2.
${{P}_{2}}+a{{P}_{1}}+2b=0$
Substituting the value of ${{P}_{1}}$ and “a” we get
$\begin{align}
& {{P}_{2}}+\left( 2 \right)\left( -2 \right)+2b=0 \\
& \Rightarrow {{P}_{2}}-4+2b=0 \\
\end{align}$
Transposing -4+2b to RHS we get
${{P}_{2}}=4-2b$
Put n = 3 in the recurrence ${{P}_{n}}+a{{P}_{n-1}}+b{{P}_{n-2}}=0$, we get
${{P}_{3}}+\left( 2 \right){{P}_{2}}+b{{P}_{1}}=0$
But ${{P}_{3}}={{\alpha }^{3}}+{{\beta }^{3}}=-56$
Substituting the value of ${{P}_{1}},{{P}_{2}}$ and ${{P}_{3}}$ we get
$\begin{align}
& -56+\left( 2 \right)\left( 4-2b \right)+b\left( -2 \right)=0 \\
& \Rightarrow -56+8-4b-2b=0 \\
& \Rightarrow -48-6b=0 \\
\end{align}$
Adding 6b on both sides, we get
$\begin{align}
& -48-6b+6b=0+6b \\
& \Rightarrow 6b=-48 \\
\end{align}$
Dividing both sides by 6, we get
$\begin{align}
& \dfrac{6b}{6}=\dfrac{-48}{6}=-8 \\
& \Rightarrow b=-8 \\
\end{align}$
Hence b = -8 and a = 2
Hence the quadratic expression is ${{x}^{2}}+ax+b={{x}^{2}}+2x-8$
[2] The derivation of the Newton's method is a direct result of the observation that if $\alpha $ is a root of quadratic expression ${{x}^{2}}+ax+b$ then ${{\alpha }^{n}}+a{{\alpha }^{n-1}}+b{{\alpha }^{n-2}}=0\forall n\ge 3$. Write a similar expression for $\beta $ and add the two equations to get the above recursive relation.
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