Question

The $\alpha ,\beta $ are real numbers such that ${{\alpha }^{3}}-3{{\alpha }^{2}}+5\alpha -17=0$ and ${{\beta }^{3}}-3{{\beta }^{2}}+5\beta +11=0$. What is the value of $\alpha +\beta $?

Answer 1

Hint: We first assume the value of the $\alpha +\beta =\gamma $. We express the equation ${{\alpha }^{3}}-3{{\alpha }^{2}}+5\alpha -17=0$ with respect to $\beta $ and get the cubic equation. We equate the coefficients of the new equation with ${{\beta }^{3}}-3{{\beta }^{2}}+5\beta +11=0$ and get the value of $\alpha ,\beta $. Then we find the value of $\alpha +\beta $.

Complete step by step answer:
We assume that the value of $\alpha +\beta =\gamma $ is real where $\alpha ,\beta $ are real numbers themselves. We get $\alpha =\gamma -\beta $. We replace the values in the equation of ${{\alpha }^{3}}-3{{\alpha }^{2}}+5\alpha -17=0$.
\[{{\alpha }^{3}}-3{{\alpha }^{2}}+5\alpha -17=0 \\
\Rightarrow {{\left( \gamma -\beta \right)}^{3}}-3{{\left( \gamma -\beta \right)}^{2}}+5\left( \gamma -\beta \right)-17=0 \]

We simplify the cubic equation
\[{{\left( \gamma -\beta \right)}^{3}}-3{{\left( \gamma -\beta \right)}^{2}}+5\left( \gamma -\beta \right)-17=0 \\
\Rightarrow {{\gamma }^{3}}-3{{\gamma }^{2}}\beta +3\gamma {{\beta }^{2}}-{{\beta }^{3}}-3{{\gamma }^{2}}-3{{\beta }^{2}}+6\beta \gamma +5\gamma -5\beta -17=0 \\
\Rightarrow {{\beta }^{3}}+\left( 3-3\gamma \right){{\beta }^{2}}+\left( 3{{\gamma }^{2}}-6\gamma +5 \right)\beta +\left( -{{\gamma }^{3}}-5\gamma +17+3{{\gamma }^{2}} \right)=0 \]

We now equate the coefficients of the equation ${{\beta }^{3}}-3{{\beta }^{2}}+5\beta +11=0$ with \[{{\beta }^{3}}+\left( 3-3\gamma \right){{\beta }^{2}}+\left( 3{{\gamma }^{2}}-6\gamma +5 \right)\beta +\left( -{{\gamma }^{3}}-5\gamma +17+3{{\gamma }^{2}} \right)=0\].
We get \[\left( 3-3\gamma \right)=-3;\beta =5;\left( -{{\gamma }^{3}}-5\gamma +17+3{{\gamma }^{2}} \right)=11\].

From the first two equations we got \[\left( 3-3\gamma \right)=-3\Rightarrow \gamma =\dfrac{3+3}{3}=2\] and \[\beta =5\]. We now just need to confirm that \[\gamma =2\] satisfies \[\left( -{{\gamma }^{3}}-5\gamma +17+3{{\gamma }^{2}} \right)=11\].
We get
\[\left( -{{2}^{3}}-5\times 2+17+3\times {{2}^{2}} \right) =-8-10+17+12 \\
\Rightarrow \left( -{{2}^{3}}-5\times 2+17+3\times {{2}^{2}} \right) =11 \\ \]
So, the values \[\gamma =2\] and \[\beta =5\] are verified. The value of $\gamma $ is $\gamma =\alpha +\beta =2+5=7$.

Therefore, the value of $\alpha +\beta $ is 7.

Note: We can also find the value of $\beta $ and express the equation ${{\beta }^{3}}-3{{\beta }^{2}}+5\beta +11=0$ with respect to $\alpha $ and get the cubic equation. In any case the equality of the coefficients has to satisfy the summation of the variables.