Question

 For the following, find a quadratic polynomial whose sum and product respectively of the zeroes are as given. Also, find the zeroes of these polynomials by factorisation.
\[\dfrac{-3}{2\sqrt{5}},\dfrac{-1}{2}\]

Answer 1

Hint: In this question, we need to find the equation of the quadratic polynomial by using the relation between the roots of the polynomial and the coefficients of the polynomial. Then by substituting the respective values of the coefficients in the general form we get the quadratic polynomial. Now, by factoring the polynomial and simplifying it further we get the zeroes.

Complete step by step answer:
QUADRATIC POLYNOMIAL: A quadratic polynomial is a polynomial having degree two.
The general form of a quadratic polynomial is given by \[a{{x}^{2}}+bx+c=0\]
ZERO OF A POLYNOMIAL: A real number \[\alpha \] is a zero of the polynomial p(x), if and only if \[p\left( \alpha \right)=0\]
Now, let us assume the zeroes of the quadratic polynomial as
\[\alpha ,\beta \]
Now, from the quadratic polynomial \[a{{x}^{2}}+bx+c=0\] we have,
Sum of the roots of the quadratic polynomial is given by
\[\alpha +\beta =\dfrac{-b}{a}\]
Now, product of all the roots is given by
\[\alpha \beta =\dfrac{c}{a}\]
Now, as already given in the question
\[\Rightarrow \alpha +\beta =\dfrac{-3}{2\sqrt{5}}\]
\[\Rightarrow \alpha \beta =\dfrac{-1}{2}\]
Now, on comparing these values with the formulae above we get,
\[\Rightarrow \dfrac{-b}{a}=\dfrac{-3}{2\sqrt{5}}\]
Now, on writing b in terms of a we get,
\[\therefore b=\dfrac{3a}{2\sqrt{5}}\]
Now, from the product of the roots formula we get,
\[\Rightarrow \dfrac{c}{a}=\dfrac{-1}{2}\]
Now, on again writing c in terms of a we get,
\[\therefore c=\dfrac{-a}{2}\]
Now, let us substitute back these values of b and c in the general form of quadratic polynomial.
\[\Rightarrow a{{x}^{2}}+bx+c=0\]
Now, on substituting the corresponding values we get,
\[\Rightarrow a{{x}^{2}}+\dfrac{3a}{2\sqrt{5}}x+\dfrac{-a}{2}=0\]
Let us now multiply with \[2\sqrt{5}\] on both the sides.
\[\Rightarrow 2\sqrt{5}a{{x}^{2}}+3ax-\sqrt{5}a=0\]
Let us now divide with a on both sides and rewrite the above equation.
\[\Rightarrow 2\sqrt{5}{{x}^{2}}+3x-\sqrt{5}=0\]
Now, let us multiply with \[\sqrt{5}\] to simplify it further then we get,
\[\therefore 10{{x}^{2}}+3\sqrt{5}x-5=0\]
Now, let us find the zeroes of the above quadratic polynomial by factorisation.
\[\Rightarrow 10{{x}^{2}}+3\sqrt{5}x-5=0\]
Now, the above quadratic equation can also be written as
\[\Rightarrow 10{{x}^{2}}+5\sqrt{5}x-2\sqrt{5}x-5=0\]
Now, let us take 5 and x common in the first two terms and \[\sqrt{5}\] in the last two terms
\[\Rightarrow 5x\left( 2x+\sqrt{5} \right)-\sqrt{5}\left( 2x+\sqrt{5} \right)=0\]
Now, this can be further factorised as
\[\Rightarrow \left( 5x-\sqrt{5} \right)\left( 2x+\sqrt{5} \right)=0\]
Now, this can be further simplified as
\[\Rightarrow 5x-\sqrt{5}=0\]
Now, on rearranging the terms we get,
\[\begin{align}
  & \Rightarrow x=\dfrac{\sqrt{5}}{5} \\
 & \therefore x=\dfrac{1}{\sqrt{5}} \\
\end{align}\]
Now, on considering the other possible case we get,
\[\Rightarrow 2x+\sqrt{5}=0\]
Now, on rearranging the terms we get,
\[\therefore x=\dfrac{-\sqrt{5}}{2}\]
Hence, the zeroes of the quadratic polynomial formed are
\[\dfrac{1}{\sqrt{5}},\dfrac{-\sqrt{5}}{2}\]

Note:
Instead of directly writing the values of b and c in terms of a from the relation between the roots of the quadratic polynomial and their coefficients we can find the value of the zeroes there itself and substitute them in the respective formula to get the coefficients.
While finding the equation of the quadratic polynomial we need to multiply with appropriate terms such that there should not be any square root term in the denominator and in the first coefficient. So, when we factorise it will be easy to get the common terms and simplify it further.
It is important to note that while substituting the respective values and simplifying we should not neglect any of the terms because it will change the equation accordingly and so the result.