Question

Find the quadratic polynomial whose zeroes are $ {{3}^{0}}\And {{2}^{-1}} $

Answer 1

Hint: We had to know the quadratic equation whose roots are given, so by using the middle term splitting method in a reverse way we will get the quadratic polynomial. Other than this we can also find it with the factor form of quadratic polynomials.

Complete step by step answer:
Moving ahead with the solution in a stepwise manner,
We know that the quadratic equation has two roots which are given as $ {{3}^{0}} $ and $ {{2}^{-1}} $ . As we know that any number where power is zero is 1 i.e. we can write $ {{3}^{0}} $ equal to 1. So we will get;
 $ {{3}^{0}}=1 $ , in the similar way if we try to reduce the other root of quadratic polynomial written in exponential form to numerical form we will get;
 $ {{2}^{-1}}=\dfrac{1}{2} $ . So we can say that we have two roots which are $ 1 $ and $ \dfrac{1}{2} $ .
By the splitting middle term method we know that if roots of any quadratic polynomial are $ \alpha $ and $ \beta $ then we can write quadratic polynomial in the form of sum and multiplication of roots as $ {{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta =0 $ . In which $ \alpha $ is one of the two roots of quadratic and the other root is $ \beta $ .
In the similar way if we write the roots then we will get the quadratic polynomial, so as according to question two roots are $ {{3}^{0}}\And {{2}^{-1}} $ so we can write them as $ 1 $ and $ \dfrac{1}{2} $ , one of which is $ \alpha $ and other is $ \beta $ . Let $ 1=\alpha $ and $ \dfrac{1}{2}=\beta $ , then according to middle term splitting method quadratic equation will be;
 $ \begin{align}
  & {{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta =0 \\
 & {{x}^{2}}-\left( 1+\dfrac{1}{2} \right)x+\left( 1 \right)\left( \dfrac{1}{2} \right)=0 \\
\end{align} $
On further simplifying it we will get;
 $ {{x}^{2}}-\left( \dfrac{3}{2} \right)x+\left( \dfrac{1}{2} \right)=0 $
On talking LCM we will get;
 $ \dfrac{2{{x}^{2}}-3x+1}{2}=0 $
By cross multiplying; we will get;
 $ 2{{x}^{2}}-3x+1=0 $
So we got the quadratic equation $ 2{{x}^{2}}-3x+1=0 $ whose roots are $ {{3}^{0}}\And {{2}^{-1}} $ .
Hence the answer is $ 2{{x}^{2}}-3x+1=0 $ .

Note: Since the roots are given in the exponential form, we reduced them to simpler rational form with only motive to have a simple calculation. Moreover rather than writing $ 2{{x}^{2}}-3x+1=0 $ as answer, we can also write $ {{x}^{2}}-\left( \dfrac{3}{2} \right)x+\left( \dfrac{1}{2} \right)=0 $ both are correct.