Question

If \[\alpha \]and \[\beta \]are the roots of \[\begin{array}{*{20}{c}}{{x^2} - 3x + 1}& = &0\end{array}\], then the equation whose roots are \[\left( {\frac{1}{{\alpha - 2}},\frac{1}{{\beta - 2}}} \right)\].
A. \[\begin{array}{*{20}{c}}{{x^2} + x - 1}& = &0\end{array}\]
B. \[\begin{array}{*{20}{c}}{{x^2} + x + 1}& = &0\end{array}\]
C. \[\begin{array}{*{20}{c}}{{x^2} - x - 1}& = &0\end{array}\]
D. None of these

Answer 1

Hint: In this question, we have given the roots of the unknown equation and we have to determine the unknown quadratic equation. First of all, we will determine the sum and the product of the roots of the given quadratic equation. After that, we will determine the sum and product of the roots of the unknown equation. And then we use the general form of the quadratic equation to determine the unknown equation. Hence, we will get a suitable answer.

Formula used:
For sum of the roots:
\[\begin{array}{*{20}{c}}{\alpha + \beta }& = &{ - \frac{b}{a}}\end{array}\]
For product of the roots:
 \[\begin{array}{*{20}{c}}{\alpha \beta }& = &{\frac{c}{a}}\end{array}\]
Quadratic equation if roots given:
\[\begin{array}{*{20}{c}}{{x^2} - \left( {\alpha + \beta } \right)x + \alpha \beta }& = &0\end{array}\]

Complete step by step solution:
According to the given question, we have given the equation whose roots are \[\alpha \]and \[\beta \] and we have also given the roots of the unknown equation that we will have to determine. Therefore,
\[ \Rightarrow \begin{array}{*{20}{c}}{{x^2} - 3x + 1}& = &0\end{array}\]
Now sum and the product of the roots of the above equation will be
\[ \Rightarrow \begin{array}{*{20}{c}}{\alpha + \beta }& = &3\end{array}\]
And
\[ \Rightarrow \begin{array}{*{20}{c}}{\alpha \beta }& = &1\end{array}\]
Now we will determine the sum of the roots of the unknown equation. Therefore, we can write,
\[ \Rightarrow \frac{1}{{\alpha - 2}} + \frac{1}{{\beta - 2}}\]
\[ \Rightarrow \frac{{\beta - 2 + \alpha - 2}}{{\left( {\alpha - 2} \right)\left( {\beta - 2} \right)}}\]
\[ \Rightarrow \frac{{\alpha + \beta - 4}}{{\alpha \beta - 2\left( {\alpha + \beta } \right) + 4}}\]
Now put the value of \[\alpha + \beta \]and \[\alpha \beta \]in the above expression. Therefore, we will get
\[ \Rightarrow \frac{{3 - 4}}{{1 - 2\left( 3 \right) + 4}}\]
\[ \Rightarrow 1\]
And then we will determine the product of the roots of the given equation. Therefore, we can get
\[ \Rightarrow \frac{1}{{\alpha - 2}} \times \frac{1}{{\beta - 2}}\]
\[ \Rightarrow \frac{1}{{\alpha \beta - 2\left( {\alpha + \beta } \right) + 4}}\]
Now we will put the value of the \[\alpha + \beta \]and \[\alpha \beta \]in the above expression. Therefore, we will get
\[ \Rightarrow \frac{1}{{1 - 2\left( 3 \right) + 4}}\]
\[ \Rightarrow - 1\]
Now for the unknown equation, we will apply
\[ \Rightarrow \begin{array}{*{20}{c}}{{x^2} - \left( {\alpha + \beta } \right)x + \alpha \beta }& = &0\end{array}\]
Now we have the sum and the product of the roots of the unknown equations such as \[1\]and \[ - 1\]. Therefore, we will get
\[ \Rightarrow \begin{array}{*{20}{c}}{{x^2} - x - 1}& = &0\end{array}\]

Therefore, the correct option is C.

Note: In this question, the first point is to keep in mind that to determine the unknown equation, we will use the sum and the product of the roots of the unknown equation. A quadratic problem is one that involves a variable multiplied by itself and the squares operation. It is important to understand that in the equation $ax^2+bx+c$, the sum of the equation's roots equals a and the product equals b. The product of a negative number and a positive number must be negative.