Question

If $c \ne 0$ and the equation $\dfrac{p}{{2x}} = \dfrac{a}{{\left( {x + c} \right)}} + \dfrac{b}{{(x - c)}}$ has $2$ equal roots, then $p$ can be
$1)a + b $
$2)a - b $
$3){\left( {\sqrt {a \pm b} } \right)^2}$
4) None of these

Answer 1

Hint: According to the question, we are given an equation which has two equal rots. A quadratic equation is in the form of $a{x^2} + bx + c = 0$. Where, $a,b,c$ are called as coefficients. A quadratic equation will have two equal roots when the value of the discriminant is zero. The discriminant is a value that depends on the coefficients. Therefore, we can rearrange the equation accordingly. Then, we can equate the discriminant value to $0$ to find the value of $p$

Complete answer:
The given equation is
$\dfrac{p}{{2x}} = \dfrac{a}{{\left( {x + c} \right)}} + \dfrac{b}{{(x - c)}}$
It can be rearranged and written as,
$\dfrac{p}{{2x}} = \dfrac{{\left( {a + b} \right)x + c(b - a)}}{{{x^2} - {c^2}}}$
By cross-multiplying, we get
$p({x^2} - {c^2}) = 2\left( {a + b} \right){x^2} - 2xc(b - a)$
This can also be written as,
$(2a + 2b - p){x^2} - 2c(a - b)x - p{c^2} = 0.......(1)$
This is in the form of $a{x^2} + bx + c = 0$
So, for (1), we have,
$\eqalign{
  & a = (2a + 2b - p) \cr
  & b = 2c(a - b) \cr
  & c = p{c^2} \cr} $
For the equation to have two equal roots, the discriminant should be zero.
The discriminant is given by,
${b^2} - 4ac = 0$
We now equate the discriminant of (1) to zero.
$ \Rightarrow {(2ca - 2b)^2} - 4(2a + 2b - p)(p{c^2}) = 0$
Now, let us multiply and subtract accordingly.
$ \Rightarrow {(2)^2}{(ac - bc)^2} - 4(2a{c^2}p + 2b{c^2}p - {p^2}{c^2}) = 0$
Let us group the terms. The terms that are in multiplication in the LHS can be transferred to the RHS and it becomes zero.
$\eqalign{
  & \Rightarrow (4c){(a - b)^2} - 4p{c^2}(2a + 2b - p) = 0 \cr
  & \Rightarrow {(a - b)^2} - p(2a + 2b - p) = 0 \cr} $
We will separate and rearrange the terms
$\eqalign{
  & \Rightarrow {(a - b)^2} + {p^2} - 2p(a + b) = 0 \cr
  & \Rightarrow {\left[ {p - (a + b)} \right]^2} = {\left( {a + b} \right)^2} - {(a - b)^2} \cr
  & \Rightarrow {p^2} = 4ab + {(a + b)^2} \cr} $
Removing square on both sides,
$\eqalign{
  & \Rightarrow {p^2} = 4ab + {(a + b)^2} \cr
  & \Rightarrow p = a + b \pm \sqrt {ab} \cr
  & \Rightarrow p = {\left( {\sqrt {a + b} } \right)^2} \cr} $
Hence, option (3) is the correct answer.

Note:
We know that a quadratic equation will have two roots.
While removing the square, do not assume that it is going to be a positive root. Use $ \pm $ to distinguish the two roots.
While simplifying there are many squares, make sure you use the proper simplification process. The condition is already given, so equate the discriminant to zero.