Question

If \[{x^2} + ax + bc = 0\] and \[{x^2} + bx + ca = 0\] (\[a \ne b\] and \[c \ne 0\]) have a common root, then \[a + b + c = \]
A.\[0\]
B.\[1\]
C.\[ab + bc + ca\]
D.\[3abc\]

Answer 1

Hint: Here, we have to find the sum of the roots. First, we will use the condition if two roots have a common root, to find the common root. Then by using the condition we will find the sum of the roots. A quadratic equation is an equation of a variable with the highest degree is 2.
Formula Used:
The difference of the square of two numbers is given by \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]

Complete step-by-step answer:
Let us consider two quadratic equations \[{a_1}{x^2} + {b_1}x + {c_1} = 0\] and \[{a_2}{x^2} + {b_2}x + {c_2} = 0\].
If the two equations have a common root, then we have the condition that,
\[\dfrac{{{x^2}}}{{{b_1}{c_1} - {c_2}{b_2}}} = \dfrac{{ - x}}{{{a_1}{c_2} - {c_1}{a_2}}} = \dfrac{1}{{{a_1}{b_2} - {b_1}{a_2}}}\] ………………………..\[\left( 1 \right)\]
Now, consider the equation \[{x^2} + ax + bc = 0\] and \[{x^2} + bx + ca = 0\].
We will now use the condition in equation \[\left( 1 \right)\] for the given equations.
Now, Considering the last two terms
\[\dfrac{{ - x}}{{{a_1}{c_2} - {c_1}{a_2}}} = \dfrac{1}{{{a_1}{b_2} - {b_1}{a_2}}}\]
By substituting the values of the coefficients and the constant term, we get
\[ \Rightarrow \dfrac{{ - x}}{{1 \cdot ca - bc \cdot 1}} = \dfrac{1}{{1 \cdot b - a \cdot 1}}\]
By taking out the common terms, we get
\[ \Rightarrow \dfrac{{ - x}}{{ca - bc}} = \dfrac{1}{{b - a}}\]
By cancelling the terms, we get
\[ \Rightarrow \dfrac{{ - x}}{{c\left( {a - b} \right)}} = \dfrac{1}{{\left( { - 1} \right)\left( {a - b} \right)}}\]
\[ \Rightarrow \dfrac{x}{c} = 1\]
By cross- multiplying the equation, we get
\[ \Rightarrow x = c\] ………………………………………………………………………………..\[\left( 2 \right)\]
Now, considering the first two terms, we get
\[\dfrac{{{x^2}}}{{{b_1}{c_1} - {c_2}{b_2}}} = \dfrac{{ - x}}{{{a_1}{c_2} - {c_1}{a_2}}}\]
By cancelling both the terms, we get
\[ \Rightarrow \dfrac{x}{{{b_1}{c_1} - {c_2}{b_2}}} = \dfrac{{ - 1}}{{{a_1}{c_2} - {c_1}{a_2}}}\]
By substituting the values of the coefficients and the constant term, we get
\[ \Rightarrow \dfrac{x}{{a \cdot ac - bc \cdot b}} = \dfrac{{ - 1}}{{1 \cdot ca - bc \cdot 1}}\]
By multiplying the terms, we get
\[ \Rightarrow \dfrac{x}{{{a^2}c - {b^2}c}} = \dfrac{{ - 1}}{{ca - bc}}\]
By taking out the common terms, we get
\[ \Rightarrow \dfrac{x}{{c\left( {{a^2} - {b^2}} \right)}} = \dfrac{{ - 1}}{{c\left( {a - b} \right)}}\]
By substituting equation \[\left( 2 \right)\] in the above equation, we get
\[ \Rightarrow \dfrac{c}{{\left( {{a^2} - {b^2}} \right)}} = \dfrac{{ - 1}}{{\left( {a - b} \right)}}\]
Using the algebraic identity \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\], we get
\[ \Rightarrow \dfrac{c}{{\left( {a + b} \right)\left( {a - b} \right)}} = \dfrac{{ - 1}}{{\left( {a - b} \right)}}\]
By cancelling out the common terms, we get
\[ \Rightarrow \dfrac{c}{{\left( {a + b} \right)}} = - 1\]
By cross-multiplying, we get
\[ \Rightarrow c = - 1\left( {a + b} \right)\]
\[ \Rightarrow c = - a - b\]
By rewriting the equation, we get
\[ \Rightarrow a + b + c = 0\]
Therefore, if \[{x^2} + ax + bc = 0\] and \[{x^2} + bx + ca = 0\] have a common root, then \[a + b + c = 0\].
Thus, option (A) is the correct answer.

Note: A quadratic equation is an equation, which has the highest degree of variable as 2 and has two solutions. We should remember the use of the condition if the quadratic equations have a common root to find the value of the common root. We also have a condition that if the two quadratic equations have a common root, then the sum of the roots is always zero whatever be the quadratic equation.