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Question: If \[{x^2} + ax + bc = 0\] and \[{x^2} + bx + ca = 0\] (\[a \ne b\] and \[c \ne 0\]) have a common r...

If x2+ax+bc=0{x^2} + ax + bc = 0 and x2+bx+ca=0{x^2} + bx + ca = 0 (aba \ne b and c0c \ne 0) have a common root, then a+b+c=a + b + c =
A.00
B.11
C.ab+bc+caab + bc + ca
D.3abc3abc

Explanation

Solution

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 a2b2=(a+b)(ab){a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)

Complete step-by-step answer:
Let us consider two quadratic equations a1x2+b1x+c1=0{a_1}{x^2} + {b_1}x + {c_1} = 0 and a2x2+b2x+c2=0{a_2}{x^2} + {b_2}x + {c_2} = 0.
If the two equations have a common root, then we have the condition that,
x2b1c1c2b2=xa1c2c1a2=1a1b2b1a2\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}}} ………………………..(1)\left( 1 \right)
Now, consider the equation x2+ax+bc=0{x^2} + ax + bc = 0 and x2+bx+ca=0{x^2} + bx + ca = 0.
We will now use the condition in equation (1)\left( 1 \right) for the given equations.
Now, Considering the last two terms
xa1c2c1a2=1a1b2b1a2\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
x1cabc1=11ba1\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
xcabc=1ba\Rightarrow \dfrac{{ - x}}{{ca - bc}} = \dfrac{1}{{b - a}}
By cancelling the terms, we get
xc(ab)=1(1)(ab)\Rightarrow \dfrac{{ - x}}{{c\left( {a - b} \right)}} = \dfrac{1}{{\left( { - 1} \right)\left( {a - b} \right)}}
xc=1\Rightarrow \dfrac{x}{c} = 1
By cross- multiplying the equation, we get
x=c\Rightarrow x = c ………………………………………………………………………………..(2)\left( 2 \right)
Now, considering the first two terms, we get
x2b1c1c2b2=xa1c2c1a2\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
xb1c1c2b2=1a1c2c1a2\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
xaacbcb=11cabc1\Rightarrow \dfrac{x}{{a \cdot ac - bc \cdot b}} = \dfrac{{ - 1}}{{1 \cdot ca - bc \cdot 1}}
By multiplying the terms, we get
xa2cb2c=1cabc\Rightarrow \dfrac{x}{{{a^2}c - {b^2}c}} = \dfrac{{ - 1}}{{ca - bc}}
By taking out the common terms, we get
xc(a2b2)=1c(ab)\Rightarrow \dfrac{x}{{c\left( {{a^2} - {b^2}} \right)}} = \dfrac{{ - 1}}{{c\left( {a - b} \right)}}
By substituting equation (2)\left( 2 \right) in the above equation, we get
c(a2b2)=1(ab)\Rightarrow \dfrac{c}{{\left( {{a^2} - {b^2}} \right)}} = \dfrac{{ - 1}}{{\left( {a - b} \right)}}
Using the algebraic identity a2b2=(a+b)(ab){a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right), we get
c(a+b)(ab)=1(ab)\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
c(a+b)=1\Rightarrow \dfrac{c}{{\left( {a + b} \right)}} = - 1
By cross-multiplying, we get
c=1(a+b)\Rightarrow c = - 1\left( {a + b} \right)
c=ab\Rightarrow c = - a - b
By rewriting the equation, we get
a+b+c=0\Rightarrow a + b + c = 0
Therefore, if x2+ax+bc=0{x^2} + ax + bc = 0 and x2+bx+ca=0{x^2} + bx + ca = 0 have a common root, then a+b+c=0a + 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.