Question

Find a quadratic polynomial whose zeros are $3-\sqrt{5}\text{ and }3+\sqrt{5}$.

Answer 1

Hint:For solving this problem, we obtain the sum and product of roots individually and then place the values of variables in the generalized form to obtain the final polynomial.

Complete step-by-step answer:
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:
$f(x)=a{{x}^{2}}+bx+c,\quad a\ne 0$
If the quadratic function is set equal to zero, then the result is a quadratic equation. The solutions to the univariate equation are called the roots of the univariate function.
If we have two zeros of a quadratic equation then the polynomial could be formed by using the simplified result which could be stated as:
${{x}^{2}}-(a+b)x+ab$, where a and b are two zeroes of the equation.
Let, $a=3+\sqrt{5}$ and $b=3+\sqrt{5}$ be the roots of the required equation.
The sum of zeroes i.e. a + b:
$\begin{aligned} & a+b=3+\sqrt{5}+3-\sqrt{5} \\\ & a+b=6\ldots (1) \\\ \end{aligned}$
The product of zeroes i.e. $a\times b:$
$\begin{aligned} & a\times b=\left( 3+\sqrt{5} \right)\times \left( 3-\sqrt{5} \right) \\\ & a\times b=9-5=4\ldots (2) \\\ \end{aligned}$
Sum of roots can also be written as $\dfrac{-b}{a}=-\dfrac{6}{1}$.
Also,
The products of roots can be written as $\dfrac{c}{a}=\dfrac{4}{1}$.
Fixing a as 1, we get
$\Rightarrow a=1,b=6,c=4$
Now, putting these values in the general form.
Hence, the quadratic polynomial is ${{x}^{2}}-6x+4$.

Note: By using the quadratic formula i.e. $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, we can evaluate the values of a, b and c alternatively. Student must be careful while writing the simplified form of sum of roots and product of roots. The negative sign in the sum of roots must not be missed to avoid errors.