Answer
Verified
479.1k+ views
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Change the following sentences into negative and interrogative class 10 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE