If a, b and c are real, then both the roots of the equation $$\left( {x - b} \right)\left( {x - c} \right) + \left( {x - c} \right)\left( {x - a} \right) + \left( {x - a} \right)\left( {x - b} \right) = 0$$ are always:A. PositiveB. NegativeC. RealD. Imaginary

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Vedantu Content Team · Accepted Answer

Hint: First of all, we will expand the given equation $$\left( {x - b} \right)\left( {x - c} \right) + \left( {x - c} \right)\left( {x - a} \right) + \left( {x - a} \right)\left( {x - b} \right) = 0$$ and write it in the powers of x and the other remaining part as constant. Doing so, we will easily sort the coefficients of powers of x. Then, we will find the discriminant D of the given equation using the formula $D = {b^2} - 4ac$. After that, we will check the conditions if D = 0 or D $ \geqslant $0 or if D $ \leqslant $0. We can then find out whether any of the options is correct or not. Complete step by step answer:We are given that a, b and c are real.The given equation is $$\left( {x - b} \right)\left( {x - c} \right) + \left( {x - c} \right)\left( {x - a} \right) + \left( {x - a} \right)\left( {x - b} \right) = 0$$. We are required to find the nature of the roots of this equation.First, we will expand and simplify the given equation as$ \Rightarrow \left( {x - b} \right)\left( {x - c} \right) + \left( {x - c} \right)\left( {x - a} \right) + \left( {x - a} \right)\left( {x - b} \right) = 0 $$   \Rightarrow {x^2} - bx - cx + bc + {x^2} - cx - ax + ac + {x^2} - ax - bx + ab = 0 $Combining the coefficients of the same powers of x, we get$ \Rightarrow 3{x^2} - 2\left( {a + b + c} \right)x + (ab + bc + ac) = 0$Now, we know that the formula to calculate the discriminant D of the any algebraic equation is given by $D = {b^2} - 4ac$, where a is the coefficient of $x^2$, b is the coefficient of x and c is the constant term of the equation.Therefore, we get a = 3, b = -2 (a + b + c) and c = (ab + bc + ac) from the equation we simplified above.Substituting these values in the formula of D, we get$   \Rightarrow D = {b^2} - 4ac $$   \Rightarrow D = {\left\{ { - 2\left( {a + b + c} \right)} \right\}^2} - 4\left( 3 \right)\left( {ab + bc + ac} \right) $$   \Rightarrow D = 4{\left( {a + b + c} \right)^2} - 4(3)\left( {ab + bc + ca} \right) $we know that$\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)$$  \Rightarrow D = 4 (a^2 + b^2+ c^2+2(ab+bc+ca) -3(ab + bc + ca)) $$ \Rightarrow D = 2\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] $Here, we can see that D is the product of 2 which is a positive integer and a total of 3 perfect squares of integers which means they are positive as well.$ \Rightarrow D \geqslant 0$ We know that if1) $D = 0$, then the roots are real and equal.2) $D \geqslant 0$, then the roots are real and distinct.3) $D < 0$, then the roots are imaginary.Hence, we can say that the roots of the given equation are real.Therefore, option (C) is correct.Note: In such questions, you can get confused with the method to be used. You can use both the methods, by finding the determinant or by factorization, to calculate the roots and to determine the nature of the roots.

If a, b and c are real, then both the roots of the equation \[\left( {x - b} \right)\left( {x - c} \right) + \left( {x - c} \right)\left( {x - a} \right) + \left( {x - a} \right)\left( {x - b} \right) = 0\] are always:A. PositiveB. NegativeC. RealD. Imaginary

If a, b and c are real, then both the roots of the equation \[\left( {x - b} \right)\left( {x - c} \right) + \left( {x - c} \right)\left( {x - a} \right) + \left( {x - a} \right)\left( {x - b} \right) = 0\] are always:
A. Positive
B. Negative
C. Real
D. Imaginary