Question

If $\alpha \,and\,\beta $ are the roots of the quadratic equation ${{x}^{2}}+bx-c=0$, the equation whose roots are b and c is-
${{x}^{2}}+\alpha x-\beta =0$
\[{{x}^{2}}-\left[ \left( \alpha +\beta \right)+\alpha \beta \right]x+\alpha \beta \left( \alpha +\beta \right)=0\]
\[{{x}^{2}}+\left[ \left( \alpha +\beta \right)+\alpha \beta \right]x+\alpha \beta \left( \alpha +\beta \right)=0\]
\[{{x}^{2}}+\left[ \left( \alpha +\beta \right)+\alpha \beta \right]x-\alpha \beta \left( \alpha +\beta \right)=0\]

Answer 1

Hint: Firstly use the formulae \[p+q=\frac{-b}{a}\] And \[pq=\frac{c}{a}\] to find the sum and product of roots to get the values of ‘b’ and ‘c’. Then use the general form of the quadratic equation given by \[{{x}^{2}}+(p+q)x+pq=0\] and use ‘b’ and ‘c’ as the roots in it. Then use the values of sum and product of roots to get the final answer.
Complete step by step answer:
To solve the given question we will write the given data first,
${{x}^{2}}+bx-c=0$
To proceed further in the solution we should know the formula given below,
Formula:
If p and q are the roots of the quadratic equation \[a{{x}^{2}}+bx-c=0\] then,
\[p+q=\frac{-b}{a}\] And \[pq=\frac{c}{a}\]
As we have given that $\alpha \,and\,\beta $ are the roots of the given quadratic equation therefore by using above formula we will get,
\[\alpha \,+\beta =-b\] And \[\alpha \,\beta =-c\]
\[-\left( \alpha \,+\beta \right)=b\] And \[-\alpha \,\beta =c\] ……………………………………………………. (1)
As we have to find a quadratic equation having roots as b and c and for that we should know the general form of quadratic equation given below,
General Form:
If p and q are the roots of a quadratic equation then the quadratic equation in terms of roots can be written as,
\[{{x}^{2}}+(p+q)x+pq=0\]
By using the above general form we can write the quadratic equation having roots ‘b’ and ‘c’ as follows,
\[{{x}^{2}}+(b+c)x+bc=0\]
If we put the values of equation (1) in the above equation we will get,
\[\therefore {{x}^{2}}+\left[ -\left( \alpha \,+\beta \right)+\left( -\alpha \,\beta \right) \right]x+\left[ -\left( \alpha \,+\beta \right) \right]\left( -\alpha \,\beta \right)=0\]
If we open the brackets and simplify the above equation we will get,
\[\therefore {{x}^{2}}+\left[ -\left( \alpha \,+\beta \right)-\alpha \,\beta \right]x-\left( \alpha \,+\beta \right)\left( -\alpha \,\beta \right)\]
Further simplification in the above equation will give,
\[\therefore {{x}^{2}}+\left[ -\left( \alpha \,+\beta \right)-\alpha \,\beta \right]x+\left( \alpha \,+\beta \right)\alpha \,\beta =0\]
If we take the negative sign outside the bracket we will get,
\[\therefore {{x}^{2}}+\left( -1 \right)\left[ \left( \alpha \,+\beta \right)+\alpha \,\beta \right]x+\left( \alpha \,+\beta \right)\alpha \,\beta =0\]
Further simplification in the above equation will give,
\[\therefore {{x}^{2}}-\left[ \left( \alpha \,+\beta \right)+\alpha \,\beta \right]x+\left( \alpha \,+\beta \right)\alpha \,\beta =0\]
Therefore if $\alpha \,and\,\beta $ are the roots of the quadratic equation ${{x}^{2}}+bx-c=0$, the equation whose roots are b and c is given by \[{{x}^{2}}-\left[ \left( \alpha \,+\beta \right)+\alpha \,\beta \right]x+\left( \alpha \,+\beta \right)\alpha \,\beta =0\].
Therefore, the correct answer is option (b).

Note: Use as many as brackets as we have used in \[{{x}^{2}}+\left[ -\left( \alpha \,+\beta \right)+\left( -\alpha \,\beta \right) \right]x+\left[ -\left( \alpha \,+\beta \right) \right]\left( -\alpha \,\beta \right)=0\] to avoid confusion of signs.