Question

Find the zeros of the polynomial ${x^2} + x - p\left( {p + 1} \right)$.

Answer 1

Hint: We are given a quadratic polynomial and we need to find its zeros. A quadratic polynomial has 2 zeros $\alpha $ and $\beta $. Zero of a equation means we have to equate that equation with 0 and find the unknown variable. The formula to find the zeros of a quadratic polynomial is
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Using this, we can find the zeros of the equation ${x^2} + x - p\left( {p + 1} \right)$.

Complete step-by-step answer:
In this question, we are given a quadratic polynomial and we need to find its zeros.
A quadratic polynomial has always two zeros.
Zero of a polynomial means we need to equate the equation with zero and find the value of the variable and value of the variable should be such, such that when we put it in the given equation, we get the value 0.
Given equation: ${x^2} + x - p\left( {p + 1} \right)$
To find the zeros of the equation, we have to equate it with 0.
Therefore, equating the given equation with 0, we get
$ \Rightarrow {x^2} + x - p\left( {p + 1} \right) = 0$
Now, this is a quadratic equation and we can solve a quadratic equation using quadratic formula
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Here, $a = 1,b = 1,c = - p\left( {p + 1} \right)$.
Therefore,
\[
   \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {{1^2} - 4\left( 1 \right)\left( { - p} \right)\left( {p + 1} \right)} }}{{2\left( 1 \right)}} \\
   \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {1 + 4p\left( {p + 1} \right)} }}{2} \\
   \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {1 + 4{p^2} + 4p} }}{2} \\
   \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {4{p^2} + 4p + 1} }}{2} \\
 \]
Now, $4{p^2} + 4p + 1 = {\left( {2p + 1} \right)^2}$
Therefore,
\[ \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {{{\left( {2p + 1} \right)}^2}} }}{2}\]
Now, square and square root gets cancelled. Therefore,
\[ \Rightarrow x = \dfrac{{ - 1 \pm \left( {2p + 1} \right)}}{2}\]
\[
   \Rightarrow x = \dfrac{{ - 1 + \left( {2p + 1} \right)}}{2} \\
   \Rightarrow x = \dfrac{{ - 1 + 2p + 1}}{2} \\
   \Rightarrow x = \dfrac{{2p}}{2} \\
   \Rightarrow x = p \\
 \] and \[
   \Rightarrow x = \dfrac{{ - 1 - \left( {2p + 1} \right)}}{2} \\
   \Rightarrow x = \dfrac{{ - 1 - 2p - 1}}{2} \\
   \Rightarrow x = \dfrac{{ - 2 - 2p}}{2} \\
   \Rightarrow x = - 1 - p \\
 \]
Hence, the zeros of the equation ${x^2} + x - p\left( {p + 1} \right)$ are p and -1-p.

Note: There are two relations between the zeros of the quadratic equation and its coefficients.
Sum of zeros $ = \dfrac{{ - b}}{a}$
Product of zeros$ = \dfrac{c}{a}$
So, our equation is ${x^2} + x - p\left( {p + 1} \right)$.
Therefore, sum of zeros$ = \dfrac{{ - b}}{a} = \dfrac{{ - 1}}{1} = - 1$ and
Product of zeros $ = \dfrac{c}{a} = \dfrac{{ - p\left( {p + 1} \right)}}{1} = - p\left( {p + 1} \right)$