Question

# 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] ${{x}^{2}}+2x-8$

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 ${{x}^{2}}-\left( a+b \right)x+ab$.

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.