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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.

Complete step-by-step answer:

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.

Complete step-by-step answer:

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.

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