Question

Find the quadratic polynomial whose zeroes are 0 and 1.

Answer 1

Hint: Take the two zeroes as \[\alpha \] and \[\beta \]. Then substitute the sum of zeroes and the product of zeroes in the formula for forming quadratic polynomials, when the zeroes are given. Substitute, simplify and get the quadratic polynomial.

Complete step-by-step answer:
We have been given the zeros of quadratic polynomials as 0 and 1.
Let us take, \[\alpha \] and \[\beta \] as the zeroes.
Thus, \[\alpha =0\] and \[\beta =1\].
We know the formula for forming a quadratic polynomial if its zeroes are given, \[{{x}^{2}}\] - (sum of zeroes) x + (product of zeroes).
\[{{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta \]
We need to find the values of \[\left( \alpha +\beta \right)\] and \[\alpha \beta \] and just substitute.
\[\begin{align}
  & \alpha +\beta =0+1=1 \\
 & \alpha .\beta =0\times 1=0 \\
 & \therefore {{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta ={{x}^{2}}-1.x+0={{x}^{2}}-x \\
\end{align}\]
Thus we got the quadratic polynomial as \[\left( {{x}^{2}}-x \right)\], whose zeroes are 0 and 1.
\[\therefore \] We got the required answer.
There is an alternate method, the quadratic polynomial with roots \[\alpha ,\beta \] can be written as \[\left( x-\alpha \right),\left( x-\beta \right)\].
We said that, \[\alpha =0\] and \[\beta =1\].
Thus the polynomial with zeros 0 and 1 becomes,
\[\left( x-\alpha \right)\left( x-\beta \right)=\left( x-0 \right)\left( x-1 \right)=x\left( x-1 \right)={{x}^{2}}-x\]
\[\therefore \] The quadratic polynomial = \[{{x}^{2}}-x\].

Note: Here the zeroes of the quadratic polynomials are given. But it is not mentioned if either one is the sum of zeroes or product of zeroes. So it’s important that we consider two variables as the roots of the polynomial. Then we should get their sum and product and then substitute in the formula.