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# Find a quadratic polynomial, the sum and product of whose zeroes are $\dfrac{1}{4},-1$ respectively.  Hint: The given question is related to quadratic equations. Try to recall the formulae related to the relation between the coefficients and sum and product of the roots of a quadratic equation.

Before proceeding with the solution, we must know about the relation between the coefficients and sum and product of the roots of the quadratic equation given by $a{{x}^{2}}+bx+c=0$ .
We know, the roots of the equation $a{{x}^{2}}+bx+c=0$ are given by the quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .
Let $\alpha$ and $\beta$ be the roots of the equation. So, $\alpha =\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}$ and $\beta =\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}$. The sum of the roots is given as $\alpha +\beta =\left( \dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a} \right)+\left( \dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a} \right)=\dfrac{-2b}{2a}=\dfrac{-b}{a}$ .
So, the sum of the roots is related to the coefficients as $\alpha +\beta =\dfrac{-b}{a}$ .
The product of the roots is given as $\alpha \beta =\left( \dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a} \right)\left( \dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a} \right)=\dfrac{{{b}^{2}}-\left( {{b}^{2}}-4ac \right)}{4{{a}^{2}}}=\dfrac{c}{a}$ .
So, the product of the roots is related to the coefficients as $\alpha \beta =\dfrac{c}{a}$ .
Now, we have $a{{x}^{2}}+bx+c=0$. On dividing the equation by $a$ , we get ${{x}^{2}}+\dfrac{b}{a}x+\dfrac{c}{a}=0.....(i)$.
We have $\alpha +\beta =\dfrac{-b}{a}$ and $\alpha \beta =\dfrac{c}{a}$ . So, we can rewrite equation $(i)$ with coefficients in the form sum and product of roots as ${{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta =0.....(ii)$.
Now, coming to the question , we are given that the sum of the zeroes of a quadratic polynomial is equal to $\dfrac{1}{4}$ and the product of zeroes is equal to $-1$ . So, we can say that if $\alpha$ and $\beta$ is the roots of the equation, then $\alpha +\beta =\dfrac{1}{4}$ and $\alpha \beta =-1$ . Substituting $\alpha +\beta =\dfrac{1}{4}$ and $\alpha \beta =-1$ in equation $(ii)$ , we get ${{x}^{2}}-\dfrac{1}{4}x-1=0$ .
$\Rightarrow 4{{x}^{2}}-x-4=0$
Hence, the quadratic polynomial having sum and product of zeroes as $\dfrac{1}{4}$ and $-1$ , respectively, is given as $4{{x}^{2}}-x-4$ .

Note: The quadratic equation with coefficients in the form sum and product of roots is given as ${{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta =0$ and not ${{x}^{2}}+\left( \alpha +\beta \right)x+\alpha \beta =0$. Students often get confused and make a mistake. Such mistakes should be avoided as they can lead to wrong answers.
View Notes  Polynomial    Systems of Linear and Quadratic Equations  Polynomial Equations  Polynomial Function  Polynomial Division  Polynomial Definition  Degree of Polynomial  CBSE Class 11 Maths Chapter 5 - Complex Numbers and Quadratic Equations Formulas  