Question

If a,b,c are real, then both the roots of the equation $\left( {x - a} \right)\left( {x - b} \right) + \left( {x - b} \right)\left( {x - c} \right) + \left( {x - c} \right)(x - a) = 0$ are always
A. positive
B. negative
C. real
D. imaginary

Answer 1

Hint:
First we will simplify the above equation and check the determinant of the resultant equation to decide if the roots are real, positive, negative or imaginary. If the determinant is less than 0 then the roots will be imaginary otherwise the roots will be real.

Complete step by step solution:
According to the question the quadratic equation is given by
 $\left( {x - a} \right)\left( {x - b} \right) + \left( {x - b} \right)\left( {x - c} \right) + \left( {x - c} \right)(x - a) = 0$
Simplifying the equation further
 $ \Rightarrow ({x^2} - ax - bx + ab) + ({x^2} - bx - cx + bc) + ({x^2} - cx - ax + ca) = 0$
Further simplification gives
 $ \Rightarrow {x^2} - 2(a + b + c)x + ab + bc + ca = 0$ …(i)
Now we will check for the determinant of the equation (i)
We know that determinant of a quadratic equation ${a_1}{x^2} + {a_2}x + {a_3}$ is given by ${\left( {{a_2}} \right)^2} - 4{a_1}{a_3}$
So, determinant of the equation (i) becomes
 $ \Rightarrow 4{\left( {a + b + c} \right)^2} - 4(ab + bc + ca)$ …(ii)
Where ${a_1} = 1,\,\,{a_2} = \, - (a + b + c),\,\,{a_3} = (ab + bc + ca)$
So, simplifying the determinant in expression (ii) we get
 $ \Rightarrow 4({a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca - ab - bc - ca)$
On adding like terms, we get,
 $ \Rightarrow 4\left( {{a^2} + {b^2} + {c^2} + ab + bc + ca} \right)$
We will take the factor of 2 inside the bracket
 $ \Rightarrow 2(2{a^2} + 2{b^2} + 2{c^2} + 2ab + 2bc + 2ca)$
Then we will arrange the terms.
 $ \Rightarrow 2\left\{ {\left( {{a^2} + 2ab + {b^2}} \right) + \left( {{b^2} + 2bc + {c^2}} \right) + \left( {{a^2} + 2ac + {c^2}} \right)} \right\}$
We will simplify a bit further to get
 $ \Rightarrow 2{(a + b)^2} + 2{(b + c)^2} + 2{(a + c)^2}$ …(iii)
So, we observe that the determinant is the square terms of 3 real number so
As the expression in (iii) is greater than zero so the determinant is greater than zero
And therefore, both the roots of the equation (i) are real.

And therefore, option C is correct.

Note:
In general, everyone looks for Sri Dhar Acharya’s formula when they face any quadratic equation .So Sri Dhar Acharya’s formula goes like , if the quadratic equation is $a{x^2} + bx + c$ then the roots of the given equations are $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ so when the determinant that is the part inside the root ${b^2} - 4ac$ is greater than zero then the roots are real and when it is less than zero the roots are imaginary.