Question

The polynomial whose zeros are -5, 4 is ${{x}^{2}}+x-\alpha =0$ then $\alpha $ ?

Answer 1

Hint: A zero or a root of a polynomial basically means that the value of the polynomial function at that value of $x$ becomes $0$ . In this question we are given that the roots of the quadratic equation ${{x}^{2}}+x-\alpha =0$ are $-5$ and $4$. This means that the equation equals zero at the values of $x=-5,4$ . So, we can substitute x as -5,4 in the equation and find the value of $\alpha $ . We can find the value of $\alpha $ by putting x as -5 and then cross check by putting x as 4.

Complete step by step solution:
This type of question is based on the concept of polynomials, specifically quadratic equations and finding their roots or zeros. A polynomial may have multiple roots based on the highest power of $x$ present in it. Usually, the number of roots equals the highest power of $x$. But, all the roots may not be real, some may be imaginary. A quadratic equation is one whose highest power of $x$is $2$ . Hence they have at most $2$ real roots.
We denote any general function by $f\left( x \right)$ . Let this quadratic function be $f\left( x \right)$.
$f\left( x \right)={{x}^{2}}+x-\alpha $ .
We are given that the zeros are $-5$ and $4$. Therefore,
$\begin{align}
  & f\left( -5 \right)=0 \\
 & f\left( 4 \right)=0 \\
\end{align}$
Substituting $x=-5$ in $f\left( x \right)$ we get,
$\begin{align}
  & f\left( -5 \right)={{\left( -5 \right)}^{2}}+\left( -5 \right)-\alpha \\
 & 0=25-5-\alpha \\
 & \Rightarrow \alpha =25-5 \\
 & \Rightarrow \alpha =20 \\
\end{align}$
We can also cross check by substituting $x=4$ in $f\left( x \right)$ ,
$\begin{align}
  & f\left( 4 \right)={{\left( 4 \right)}^{2}}+\left( 4 \right)-\alpha \\
 & 0=16+4-\alpha \\
 & \Rightarrow \alpha =16+4 \\
 & \Rightarrow \alpha =20 \\
\end{align}$

Hence, the value of $\alpha =20$ .

Note: Another way to solve this problem is to write down the quadratic equation in terms of its roots. Any polynomial can be written down in terms of their roots. The general equation of a quadratic polynomial $f\left( x \right)$ whose roots are $a$ and $b$ is,
$\begin{align}
  & f\left( x \right)={{x}^{2}}-\left( sum\,of\,the\,roots \right)x+\left( product\,of\,the\,roots \right) \\
 & f\left( x \right)={{x}^{2}}-\left( a+b \right)x+\left( a\times b \right) \\
 & f\left( x \right)={{x}^{2}}-\left( a+b \right)x+ab \\
\end{align}$
We are given $f\left( x \right)={{x}^{2}}+x-\alpha ......\left( i \right)$
 zeros are,
 $\begin{align}
  & a=-5 \\
 & b=4 \\
\end{align}$
Therefore, substituting $a$ and $b$in $f\left( x \right)$ , we get,
  $\begin{align}
  & f\left( x \right)={{x}^{2}}-\left( -5+4 \right)x+\left( -5\times 4 \right) \\
 & \Rightarrow f\left( x \right)={{x}^{2}}-\left( -1 \right)x+\left( -20 \right) \\
 & \Rightarrow f\left( x \right)={{x}^{2}}+x-20\,......\left( ii \right) \\
\end{align}$
Now, comparing equations $\left( i \right)$ and $\left( ii \right)$ , we get
$\alpha =20$
Hence, we can use any two methods to solve this question and the value of $\alpha $ will be $20$ .