Questions & Answers
Question
Answers
Related Questions
Form a quadratic polynomial whose zeroes are 3 and -1.
Answer
![Verified]()
Verified
Verified
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:
\[p\left( x \right) = {x^2} - \left( {{\text{sum of zeroes}}} \right)x + \left( {{\text{product of zeroes}}} \right)\].
×
Sorry!, This page is not available for now to bookmark.
Like this
Liked
-
LIVECourses
-
Crash Courses 2021
-
Vedantu Pro
-
Vedantu Assist
-
Class 12
-
-
JEE
-
JEE Main & Advanced 2021
-
JEE Main & Advanced 2022
-
-
NEET
-
NEET 2021
-
NEET 2022
-
-
NDA
-
CBSE
-
Commerce
-
Class 11
-
Class 12
-
-
ICSE
-
Maharashtra Board
-
Class 9
-
Class 10
-
-
Micro Courses
-
-
FREEMaster Classes
-
PREMIUM1-on-1 Classes
-
Study Material
-
NCERT Solutions
-
Previous Year Papers
-
Mock Tests
-
Sample Papers
-
Reference Book Solutions
-
ICSE Solutions
-
School Syllabus
-
Revision Notes
-
Important Questions
-
Math Formula Sheets
-
-
More
Book your FREE Online counselling session
Share your contact information
Your FREE Online counselling session has been booked!
Vedantu academic counsellor will be calling you shortly for your Online Counselling session.
DONE
Please try again later
OK
NCERT Book Solutions
Reference Book Solutions
Toll Free:
1800-120-456-456
Mobile:
+91 988-660-2456
Toll Free:
1800-120-456-456
Mobile:
+91 988-660-2456
(Mon-Sun: 9am - 11pm IST)
Questions?
Chat with us
