Question

Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:
$p(y)={{y}^{2}}+\dfrac{3\sqrt{5}}{2}y-5$

Answer 1

Hint: For solving this problem, we first expand the expression given in p(y) and try to factorise by splitting the middle term method. Now, we get the zeroes of the equation. By using the property of sum of zero and product of zero, we can easily verify the relationship between coefficients and zeros.

Complete Step-by-Step solution:
In algebra, a quadratic function is a polynomial function with one or more variables in which the highest-degree term is of the second degree. A single-variable quadratic function can be stated as:
$f(x)=a{{x}^{2}}+bx+c,\quad a\ne 0$
In the above equation, the sum of zeroes is given as $\dfrac{-b}{a}$ and the product of zeroes is given as $\dfrac{c}{a}$.
Now, expanding our given expression, we get
\[\begin{align}
  & p(y)={{y}^{2}}+\dfrac{3\sqrt{5}}{2}y-5 \\
 & \Rightarrow {{y}^{2}}-\dfrac{\sqrt{5}}{2}y+2\sqrt{5}y-5=0 \\
 & \Rightarrow y\left( y-\dfrac{\sqrt{5}}{2} \right)+2\sqrt{5}\left( y-\dfrac{\sqrt{5}}{2} \right)=0 \\
 & \Rightarrow \left( y-\dfrac{\sqrt{5}}{2} \right)\left( y+2\sqrt{5} \right)=0 \\
\end{align}\]
Zeroes of the quadratic polynomial equation are: $y=\dfrac{\sqrt{5}}{2},y=-2\sqrt{5}$.
In our case, now evaluating the sum and products of zeroes by substituting $\alpha =\dfrac{\sqrt{5}}{2}\text{ and }\beta \text{=}-2\sqrt{5}$. So, mathematically it could be expressed as:
$\begin{align}
  & \alpha +\beta =\dfrac{\sqrt{5}}{2}-2\sqrt{5} \\
 & \alpha +\beta =\dfrac{\sqrt{5}-4\sqrt{5}}{2} \\
 & \alpha +\beta =\dfrac{-3\sqrt{5}}{2}\ldots (1) \\
 & \alpha \times \beta =\dfrac{\sqrt{5}}{2}\times -2\sqrt{5} \\
 & \alpha \times \beta =-5\ldots (2) \\
\end{align}$
Also, from the given equation \[p(y)={{y}^{2}}+\dfrac{3\sqrt{5}}{2}y-5\].
Now, evaluating the ratio of coefficients to verify the relationship:
$\begin{align}
  & \dfrac{-b}{a}=-\left( \dfrac{3\sqrt{5}}{2} \right) \\
 & \dfrac{-b}{a}=\dfrac{-3\sqrt{5}}{2}\ldots (3) \\
 & \dfrac{c}{a}=\dfrac{-5}{1} \\
 & \dfrac{c}{a}=-5\ldots (4) \\
\end{align}$
Therefore, from equation (1), (2), (3) and (4) we obtain $\alpha +\beta =\dfrac{-b}{a}\text{ and }\alpha \beta =\dfrac{c}{a}$.
Hence, the relationship between zeroes and coefficients is verified.

Note: Student must remember the expression for sum and product of roots of quadratic polynomial for solving this problem. For finding the zeroes of the polynomial, we can alternatively use the discriminant formula which can be expressed as: $D=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.