Questions & Answers

Question

Answers

A.Positive

B.Negative

C.Real

D.Imaginary

Answer
Verified

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

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\left[ {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right] \\

\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 if

1)$D = 0$, 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,