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Question
AnswersForm a quadratic polynomial whose zeroes are 3 and -1.
Answer
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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:
\[p\left( x \right) = {x^2} - \left( {{\text{sum of zeroes}}} \right)x + \left( {{\text{product of zeroes}}} \right)\].