Find a quadratic polynomial whose zeroes are 2 and -5 respectively.
Answer
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Hint: If Zeroes of the quadratic equation are known \[\alpha \] and \[\beta \] then the quadratic polynomial can be represented as \[{{x}^{2}}-(\alpha +\beta )x+\alpha \beta \]. Put the value of given zeroes in the formula and you will get the answer.
Complete answer:
Given the zeros of a quadratic polynomial are 2 and \[-5\]. We have to find the quadratic equation.
Consider \[\alpha \]and \[\beta \] be he zeroes of the polynomial then, \[\alpha =2\] and\[\beta =-5\]
General quadratic form is\[{{x}^{2}}-(\text{sum}\,\,\text{of}\,\,\text{Zeroes})x+\text{product}\,\,\text{of}\,\,\text{Zeroes}\].
Where, sum of roots \[=(\alpha +\beta )\]
Product of roots \[=\alpha \beta \]
Now, the quadratic equation is written as \[{{x}^{2}}-(\alpha +\beta )x+\alpha \beta ---(1)\]
We have to find the sum of the zeroes and the product of the zeroes.
Sum of zeroes is given by
Sum of zeroes \[=(\alpha +\beta )\]
Substitute the values of zeroes in the above equation
Sum of zeroes \[=(2+(-5))\]
By simplifying further we get:
Sum of zeroes \[=2-5=-3\]
Therefore, Sum of zeroes \[=-3\]
Product of zeroes is given by
Product of zeroes \[=\alpha \beta \]
Substitute the values of zeroes in the above equation
Product of zeroes \[=2(-5)\]
By simplifying further we get:
Product of zeroes \[=-(2\times 5)=-10\]
Therefore, Product of zeroes \[=-10\]
Now, on putting the above value on equation (1) we get:
\[\Rightarrow {{x}^{2}}-(-3)x+(-10)\]
On simplifying the equation we get:
\[\Rightarrow {{x}^{2}}-(-3)x-10\]
By further solving we get the final equation
\[\Rightarrow {{x}^{2}}+3x-10\]
The quadratic equation\[={{x}^{2}}+3x-10\]
Note:
Zeros polynomial values are those that, when entered into the polynomial, equal zero. The roots of a polynomial are zeros. Because the degree of the quadratic equation is also 2, it has two roots. The quadratic polynomial can be represented as \[{{x}^{2}}-(\text{sum}\,\,\text{of}\,\,\text{roots})x+\text{product}\,\,\text{of}\,\,\text{roots}=0\]. In case if you don’t remember the direct formula then also use an alternative method that is you can assume the two roots make two equations using the information given in the question. Solve to find those roots and write the quadratic equation as \[(x-{{x}_{1}})(x-{{x}_{2}})=0\] by simplifying this quadratic equation you will get the required answer as above in the solution.
Complete answer:
Given the zeros of a quadratic polynomial are 2 and \[-5\]. We have to find the quadratic equation.
Consider \[\alpha \]and \[\beta \] be he zeroes of the polynomial then, \[\alpha =2\] and\[\beta =-5\]
General quadratic form is\[{{x}^{2}}-(\text{sum}\,\,\text{of}\,\,\text{Zeroes})x+\text{product}\,\,\text{of}\,\,\text{Zeroes}\].
Where, sum of roots \[=(\alpha +\beta )\]
Product of roots \[=\alpha \beta \]
Now, the quadratic equation is written as \[{{x}^{2}}-(\alpha +\beta )x+\alpha \beta ---(1)\]
We have to find the sum of the zeroes and the product of the zeroes.
Sum of zeroes is given by
Sum of zeroes \[=(\alpha +\beta )\]
Substitute the values of zeroes in the above equation
Sum of zeroes \[=(2+(-5))\]
By simplifying further we get:
Sum of zeroes \[=2-5=-3\]
Therefore, Sum of zeroes \[=-3\]
Product of zeroes is given by
Product of zeroes \[=\alpha \beta \]
Substitute the values of zeroes in the above equation
Product of zeroes \[=2(-5)\]
By simplifying further we get:
Product of zeroes \[=-(2\times 5)=-10\]
Therefore, Product of zeroes \[=-10\]
Now, on putting the above value on equation (1) we get:
\[\Rightarrow {{x}^{2}}-(-3)x+(-10)\]
On simplifying the equation we get:
\[\Rightarrow {{x}^{2}}-(-3)x-10\]
By further solving we get the final equation
\[\Rightarrow {{x}^{2}}+3x-10\]
The quadratic equation\[={{x}^{2}}+3x-10\]
Note:
Zeros polynomial values are those that, when entered into the polynomial, equal zero. The roots of a polynomial are zeros. Because the degree of the quadratic equation is also 2, it has two roots. The quadratic polynomial can be represented as \[{{x}^{2}}-(\text{sum}\,\,\text{of}\,\,\text{roots})x+\text{product}\,\,\text{of}\,\,\text{roots}=0\]. In case if you don’t remember the direct formula then also use an alternative method that is you can assume the two roots make two equations using the information given in the question. Solve to find those roots and write the quadratic equation as \[(x-{{x}_{1}})(x-{{x}_{2}})=0\] by simplifying this quadratic equation you will get the required answer as above in the solution.
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