 Questions & Answers    Question Answers

# 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
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:  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 ${{x}^{2}}+ax+b={{x}^{2}}+2x-8$
 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.
Bookmark added to your notes.
View Notes
Beta Function  Beta Distribution  Beta Decay  Radioactivity Beta Decay  Alpha Decay  Alpha Particle Mass  Radioactivity Alpha Decay  Reaction Due to Alpha Hydrogen  Use of If Then Statements in Mathematical Reasoning  CBSE Class 10 Maths Chapter 4 - Quadratic Equations Formula  