Answer
424.8k+ views
Hint:Here first we will assume the given zeros to be \[\alpha \] and \[\beta \] then we will find the sum and product of the zeroes and substitute the respective values to get the desired quadratic equation.
Formula used:For a standard quadratic equation \[a{x^2} + bx + c = 0\] and if \[\alpha \] and \[\beta \] are the roots/zeros of the equation then
The sum of the zeroes is given by:-
\[\alpha + \beta = \dfrac{{ - b}}{a}\]
The product of the zeroes is given by:-
\[\alpha \beta = \dfrac{c}{a}\]
Complete step-by-step answer:
The given zeros are:-
3 and -1
Let \[\alpha = 3\] and \[\beta = - 1\]
Now since we know that for standard quadratic equation \[a{x^2} + bx + c = 0\] and if \[\alpha \] and \[\beta \] are the roots/zeroes of the equation then
The sum of the zeroes is given by:-
\[\alpha + \beta = \dfrac{{ - b}}{a}\]
Hence putting in the known values we get:-
\[3 - 1 = \dfrac{{ - b}}{a}\]
Solving it further we get:-
\[\dfrac{2}{1} = \dfrac{{ - b}}{a}\]
Comparing the values we get:-
\[
- b = 2 \\
\Rightarrow b = - 2 \\
\]
And, \[a = 1\]
Now we also know that the product of the zeroes is given by:-
\[\alpha \beta = \dfrac{c}{a}\]
Now putting in the known values we get:-
\[3\left( { - 1} \right) = \dfrac{c}{a}\]
Simplifying it further we get:-
\[\dfrac{{ - 3}}{1} = \dfrac{c}{a}\]
Comparing the values we get:-
\[c = - 3\] and \[a = 1\]
Putting the values of a, b and c in the standard quadratic equation we get:-
The standard quadratic equation is given by:
\[a{x^2} + bx + c = 0\]
Putting in the values of a, b, c we get:-
\[\left( 1 \right){x^2} + \left( { - 2} \right)x + \left( { - 3} \right) = 0\]
Simplifying it further we get:-
\[{x^2} - 2x - 3 = 0\]
Hence the quadratic equation is:
\[{x^2} - 2x - 3 = 0\]
The quadratic polynomial is:
\[{x^2} - 2x - 3\].
Note:The quadratic polynomial can also be written directly by using the sum and product of zeroes as:
\[p\left( x \right) = {x^2} - \left( {{\text{sum of zeroes}}} \right)x + \left( {{\text{product of zeroes}}} \right)\].
Formula used:For a standard quadratic equation \[a{x^2} + bx + c = 0\] and if \[\alpha \] and \[\beta \] are the roots/zeros of the equation then
The sum of the zeroes is given by:-
\[\alpha + \beta = \dfrac{{ - b}}{a}\]
The product of the zeroes is given by:-
\[\alpha \beta = \dfrac{c}{a}\]
Complete step-by-step answer:
The given zeros are:-
3 and -1
Let \[\alpha = 3\] and \[\beta = - 1\]
Now since we know that for standard quadratic equation \[a{x^2} + bx + c = 0\] and if \[\alpha \] and \[\beta \] are the roots/zeroes of the equation then
The sum of the zeroes is given by:-
\[\alpha + \beta = \dfrac{{ - b}}{a}\]
Hence putting in the known values we get:-
\[3 - 1 = \dfrac{{ - b}}{a}\]
Solving it further we get:-
\[\dfrac{2}{1} = \dfrac{{ - b}}{a}\]
Comparing the values we get:-
\[
- b = 2 \\
\Rightarrow b = - 2 \\
\]
And, \[a = 1\]
Now we also know that the product of the zeroes is given by:-
\[\alpha \beta = \dfrac{c}{a}\]
Now putting in the known values we get:-
\[3\left( { - 1} \right) = \dfrac{c}{a}\]
Simplifying it further we get:-
\[\dfrac{{ - 3}}{1} = \dfrac{c}{a}\]
Comparing the values we get:-
\[c = - 3\] and \[a = 1\]
Putting the values of a, b and c in the standard quadratic equation we get:-
The standard quadratic equation is given by:
\[a{x^2} + bx + c = 0\]
Putting in the values of a, b, c we get:-
\[\left( 1 \right){x^2} + \left( { - 2} \right)x + \left( { - 3} \right) = 0\]
Simplifying it further we get:-
\[{x^2} - 2x - 3 = 0\]
Hence the quadratic equation is:
\[{x^2} - 2x - 3 = 0\]
The quadratic polynomial is:
\[{x^2} - 2x - 3\].
Note:The quadratic polynomial can also be written directly by using the sum and product of zeroes as:
\[p\left( x \right) = {x^2} - \left( {{\text{sum of zeroes}}} \right)x + \left( {{\text{product of zeroes}}} \right)\].
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)