If the equations ${x^2} + ax + bc = 0\,and\,{x^2} + bx + ca = 0$ have a common root and if a, b and c are non-zero distinct real numbers, then their roots satisfy the equation:
A. ${x^2} + x + abc = 0\,\,$
B. ${x^2} - \left( {a + b} \right)x + ab = 0\,\,$
C. ${x^2} + \left( {a + b} \right)x + ab = 0\,\,$
D. ${x^2} + x + ab = 0\,\,$
E. ${x^2} + abx + abc = 0\,\,$
Answer
327.3k+ views
Hint: We have been given with two equations, to find a common root we will solve both the equations.
We will subtract the second equation from the first one,
$\left( {a - b} \right)x + c\left( {b - a} \right) = 0$
$ \Rightarrow \left( {a - b} \right)\left( {x - c} \right) = 0$
is the common root.
Thus, the roots of ${x^2} + ax + bc = 0$ are b and c and that of ${x^2} + bx + ca = 0$ are c and a.
If roots are b and a, then the equation will be,
${x^2} - \left( {a + b} \right)x + ab = 0\,$
Answer is option B
Note: In this question, we solved the equations first and then found the common root and then form the equation using the roots.
We will subtract the second equation from the first one,
$\left( {a - b} \right)x + c\left( {b - a} \right) = 0$
$ \Rightarrow \left( {a - b} \right)\left( {x - c} \right) = 0$
is the common root.
Thus, the roots of ${x^2} + ax + bc = 0$ are b and c and that of ${x^2} + bx + ca = 0$ are c and a.
If roots are b and a, then the equation will be,
${x^2} - \left( {a + b} \right)x + ab = 0\,$
Answer is option B
Note: In this question, we solved the equations first and then found the common root and then form the equation using the roots.
Last updated date: 26th May 2023
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