Question

If the sum of the root of the quadratic equation $\dfrac{1}{{x + p}} + \dfrac{1}{{x + q}} = \dfrac{1}{r}$ is zero. Show that the product of the root is $ - \dfrac{{{p^2} + {q^2}}}{2}$.

Answer 1

Hint: We can make the equation to the form of $a{x^2} + bx + c = 0$ by cross multiplication and rearranging. Then we can find the sum of the roots using the formula \[\dfrac{{ - b}}{a}\] and equate it to zero. Then we can find the product of the roots using the formula \[\dfrac{c}{a}\]. Then by making proper substitution from the 1st equation, we can eliminate r and get the required solution.

Complete step-by-step answer:
We have the equation $\dfrac{1}{{x + p}} + \dfrac{1}{{x + q}} = \dfrac{1}{r}$.
We can cross multiply the terms of LHS and take their sum.
$ \Rightarrow \dfrac{{x + q + x + p}}{{\left( {x + p} \right)\left( {x + q} \right)}} = \dfrac{1}{r}$
On simplification, we get,
 $ \Rightarrow \dfrac{{2x + q + p}}{{{x^2} + px + qx + pq}} = \dfrac{1}{r}$
On cross multiplication, we get,
$ \Rightarrow r\left( {2x + q + p} \right) = 1\left( {{x^2} + px + qx + pq} \right)$
On expanding the brackets, we get,
$ \Rightarrow 2xr + qr + pr = {x^2} + px + qx + pq$
On rearranging, we get,
$ \Rightarrow {x^2} + px + qx - 2xr + pq - qr - pr = 0$
We can make the above equation to the form $a{x^2} + bx + c = 0$
$ \Rightarrow {x^2} + \left( {p + q - 2r} \right)x + \left( {pq - qr - pr} \right) = 0$
Here,$a = 1$, $b = p + q - 2r$and $c = pq - qr - pr$
The sum of the roots is given by \[\dfrac{{ - b}}{a}\].
$ \Rightarrow S = \dfrac{{ - b}}{a}$
On substituting the values, we get,
$ \Rightarrow S = \dfrac{{ - \left( {p + q - 2r} \right)}}{1}$
It is given that the sum of the root is zero.
$ \Rightarrow - p - q + 2r = 0$
On rearranging, we get,
$ \Rightarrow 2r = p + q$
On dividing throughout with 2, we get,
$ \Rightarrow r = \dfrac{{p + q}}{2}$ … (1)
We know that the product of the roots is given by \[\dfrac{c}{a}\].
$ \Rightarrow {\text{P = }}\dfrac{c}{a}$
On substituting the values, we get,
$ \Rightarrow {\text{P = }}\dfrac{{pq - qr - pr}}{1}$
On simplification, we get,
$ \Rightarrow {\text{P = }}pq - \left( {p + q} \right)r$
On substituting value of r from equation (1), we get,
$ \Rightarrow {\text{P = }}pq - \left( {p + q} \right) \times \dfrac{{p + q}}{2}$
On rearranging, we get,
\[ \Rightarrow {\text{P = }}\dfrac{{2pq - {{\left( {p + q} \right)}^2}}}{2}\]
We can expand the square using the algebraic identity ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$, we get,
\[ \Rightarrow {\text{P = }}\dfrac{{2pq - \left( {{p^2} + {q^2} + 2pq} \right)}}{2}\]
On further calculations, we get,
\[ \Rightarrow {\text{P = }}\dfrac{{ - \left( {{p^2} + {q^2}} \right)}}{2}\]
Therefore, the product of the roots is \[\dfrac{{ - \left( {{p^2} + {q^2}} \right)}}{2}\].
Hence proved.

Note: A quadratic equation can have a maximum of 2 roots. The standard form of a quadratic equation is given by $a{x^2} + bx + c = 0$. For a quadratic equation in this form, the sum of the roots is given by \[\dfrac{{ - b}}{a}\] and the product of the root is given by \[\dfrac{c}{a}\]. We need to convert a quadratic equation to its standard form for finding the roots or values related to roots. We must eliminate the term r using equation (1) as there is no r in any of the options. We must take care of the negative signs while rearranging and simplifying.