Question

Find the quadratic polynomials whose zeroes are \[\dfrac{-2}{\sqrt{3}}\] and \[\dfrac{\sqrt{3}}{-4}\].

Answer 1

Hint: To make a quadratic equation with the help of given zeros we have a formula as \[{{x}^{2}}+(\text{sum of the zeroes})x+(\text{product of zeroes)=0}\]. Substituting the values of sum of zeroes and product of them we can easily calculate the desired quadratic equation.

Complete step-by-step answer:

We have to find a quadratic polynomial with the zeros as \[\dfrac{-2}{\sqrt{3}}\] and \[\dfrac{\sqrt{3}}{-4}\].

Now because the quadratic polynomial has degree 2so the maximum power of x in the polynomial would be 2, that is, the quadratic polynomial starts with x2.

To do so we have a formula to make a polynomial given as
\[{{x}^{2}}+(\text{sum of the zeroes})x+(\text{product of zeroes)=0}\]

Now we calculate the sum and the products of the given roots or zeros,
Sum of zeroes \[\dfrac{-2}{\sqrt{3}}\] and \[\dfrac{\sqrt{3}}{-4}\]is
\[\dfrac{-2}{\sqrt{3}}+\dfrac{\sqrt{3}}{-4}\]

Calculating this by taking LCM,
\[\begin{align}
& \Rightarrow \dfrac{-2}{\sqrt{3}}+\dfrac{\sqrt{3}}{-4}=\dfrac{8+3}{-4\sqrt{3}} \\
& \Rightarrow \dfrac{-2}{\sqrt{3}}+\dfrac{\sqrt{3}}{-4}=-\dfrac{11}{4\sqrt{3}} \\
\end{align}\]

So, we obtain the sum of zeroes as \[-\dfrac{11}{4\sqrt{3}}\]…………. (i)

Now we proceed to calculate the product of the zeroes which is given as,
\[\left( \dfrac{-2}{\sqrt{3}} \right)\left( \dfrac{\sqrt{3}}{-4} \right)\].

Solving the expression obtained above we have,
\[\begin{align}
& \Rightarrow \left( \dfrac{-2}{\sqrt{3}} \right)\left( \dfrac{\sqrt{3}}{-4} \right)=\dfrac{-2\sqrt{3}}{-4\sqrt{3}} \\
& \Rightarrow \left( \dfrac{-2}{\sqrt{3}} \right)\left( \dfrac{\sqrt{3}}{-4} \right)=\dfrac{1}{2} \\
\end{align}\]

Therefore, the product of zeroes is given by \[\dfrac{1}{2}\]………….(ii)
Using the values obtained in (i) and (ii) and substituting their value in the expression \[{{x}^{2}}+(\text{sum of the zeroes})x+(\text{product of zeroes)=0}\] we have,
\[{{x}^{2}}+(-\dfrac{11}{4\sqrt{3}})x+(\dfrac{1}{2}\text{)=0}\]

Solving the above expression,
\[\begin{align}
& \Rightarrow {{x}^{2}}-\dfrac{11}{4\sqrt{3}}x+\dfrac{1}{2}\text{=0} \\
& \Rightarrow {{x}^{2}}-\dfrac{11}{4\sqrt{3}}x+\dfrac{1}{2}\text{=0} \\
& \Rightarrow \dfrac{4\sqrt{3}{{x}^{2}}-11x+2\sqrt{3}}{4\sqrt{3}}=0 \\
\end{align}\]

Cross multiplying the above equation we get,
\[\Rightarrow 4\sqrt{3}{{x}^{2}}-11x+2\sqrt{3}=0\]
Therefore, we get our required equation as \[4\sqrt{3}{{x}^{2}}-11x+2\sqrt{3}=0\].
Hence, we obtain the quadratic polynomials whose zeroes are \[\dfrac{-2}{\sqrt{3}}\] and \[\dfrac{\sqrt{3}}{-4}\] as \[4\sqrt{3}{{x}^{2}}-11x+2\sqrt{3}=0\].

Note: A minute but huge error in the question can be putting the biggest power of x as some other integer other than 2, which is wrong because we always have the degree of quadratic equation as 2, this is why it has two zeroes.