Question

Given that the equation $kx\left( {x - 2} \right) + 6 = 0$ has real and equal roots, then the root is
$
  (a){\text{ 2}} \\
  (b){\text{ - 1}} \\
  (c){\text{ 1}} \\
  (d){\text{ }}\dfrac{1}{2} \\
 $

Answer 1

Hint: There can be two methods to solve this problem, the first be will assume the roots to be $\alpha ,\alpha $as both are equal and real, now use the relation between sum of roots which can be expressed in terms of coefficients of the given powers of the equation to get the value of roots, the second one we will explain in later end.

Complete Step-by-Step solution:
Given equation is
$kx\left( {x - 2} \right) + 6 = 0$
Now it is given that this equation has real and equal roots.
So let the roots of this quadratic equation are $\alpha ,\alpha $, where $\alpha $is real.
So first convert this equation into standard form we have,
$ \Rightarrow k{x^2} - 2kx + 6 = 0$
Now as we know that in a quadratic equation the sum of the roots is the ratio of negative times the coefficient of x to the coefficient of x2.
So in the above equation the coefficient of x is (-2k) and the coefficient of x2 is k.
Therefore sum of roots is
$ \Rightarrow \alpha + \alpha = \dfrac{{ - \left( { - 2k} \right)}}{k} = 2$
Now simplify it we have,
$ \Rightarrow 2\alpha = 2$
$ \Rightarrow \alpha = 1$.
So the root of this equation is 1.
Hence option (C) is correct.

Note: The second method will involve the concept that if a quadratic equation has real and equal roots the${b^2} - 4ac = 0$, here b, a and c are the coefficients that can be obtained by comparing the given equation with the general quadratic equation of $a{x^2} + bx + c = 0$. Through this the value of k can be obtained and then simply by using the middle term splitting concept the roots of the equation thus formed can be calculated.