Question

Find a quadratic polynomial for which the sum of zeros is 4 and product of zeros is 1.

Answer 1

Hint: Use basic concepts of quadratic equations to find the polynomial. Apply the relation between the product of zeros and sum of zeros to arrive at the solution. The formula \[{{x}^{2}}-\left( \alpha +\beta \right)x+\left( \alpha \beta \right)=0\]
Complete step by step solution:
Roots are also called x-intercepts or zeros. A quadratic function is graphically represented by a parabola with vertex located at the origin, below the x-axis, or above the x-axis. Therefore, a quadratic function may have one, two, or zero roots.
When we are asked to solve a quadratic equation, we are really being asked to find the roots. We have already seen that completing the square is a useful method to solve quadratic equations. This method can be used to derive the quadratic formula, which is used to solve quadratic equations.
The roots of a function are the x-intercepts. By definition, the y-coordinate of points lying on the x-axis is zero. Therefore, to find the roots of a quadratic function, we set f (x) = 0, and solve the equation.
Let \[\alpha ~and~\beta ~\] be the zeros or roots of a polynomial.
\[\begin{align}
  & \Rightarrow ~~\alpha +\beta =4~ \\
 & \Rightarrow ~~\alpha \beta =1~\text{ }~\text{ }~\text{ } \\
\end{align}\]
Let us denote the sum and product of zeros by S and P respectively.
\[\begin{align}
  & S=\alpha +\beta \\
 & P=\alpha \beta \\
\end{align}\]
\[\begin{align}
  & p(x)={{x}^{2}}-ax-bx+ab \\
 & \Rightarrow p(x)={{x}^{2}}-(a+b)x+ab \\
 & \Rightarrow p(x)={{x}^{2}}-Sx+P \\
\end{align}\]
Applying the conditions of the equation,
The required polynomial,
\[{{x}^{2}}-\left( \alpha +\beta \right)x+\left( \alpha \beta \right)=0\]
Substituting the values, we get
⇒ \[{{x}^{2}}-4x+1=0~\text{ }~\text{ }~\text{ }~\text{ }~\]

Note: Substitution of relevant values of the zeros must be done. It is important to remember that in the final equation we get the coefficient of the sum of zeros is negative while that of the product of roots is positive.