If \[\alpha ,\beta \] are the zeroes of the polynomials \[2{{y}^{2}}+7y+5\], then find the value of \[\alpha +\beta +\alpha \beta \].
Answer
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Hint: We are given the zeros of the polynomial \[2{{y}^{2}}+7y+5\], which are, \[\alpha ,\beta \]. So, first we have to figure the values of these zeroes and then we have to find the value of \[\alpha +\beta +\alpha \beta \]. So, in order to figure out the values of the zeroes, we will use the formula for sum of the zeroes and the product of zeroes. The given polynomial is of the form \[a{{y}^{2}}+by+c\]. So , Sum of zeroes = \[\alpha +\beta =-\dfrac{b}{a}\] and the product of zeroes = \[\alpha \beta =\dfrac{c}{a}\]. Then, we will have the values of the zeroes and we will substitute these in the expression \[\alpha +\beta +\alpha \beta \] and get the value of the expression. Hence, we will have the required value of the expression.
Complete step-by-step solution:
According to the given question, we are given the zeros of a polynomial \[2{{y}^{2}}+7y+5\] and we are asked to find the value of \[\alpha +\beta +\alpha \beta \].
The polynomial that we have is,
\[2{{y}^{2}}+7y+5\]
It is of the form, \[a{{y}^{2}}+by+c\] and so we get the corresponding values of a, b and c as,
\[a=2,b=7,c=5\]
We are given that \[\alpha ,\beta \] are the zeroes of the polynomial. So, we will first figure out the value of these zeroes and then we will find the value of the expression.
So, we have,
Sum of the zeroes = \[\alpha +\beta =-\dfrac{b}{a}\]
\[\Rightarrow \alpha +\beta =-\dfrac{7}{2}\]
Next, we have,
The product of zeroes = \[\alpha \beta =\dfrac{c}{a}\]
We get.
\[\Rightarrow \alpha \beta =\dfrac{5}{2}\]
Now, we can find the value of the given expression, that is, \[\alpha +\beta +\alpha \beta \]
So, we get,
\[\Rightarrow \left( \alpha +\beta \right)+\alpha \beta \]
We will now substitute the values in the above expression and we get,
\[\Rightarrow \left( -\dfrac{7}{2} \right)+\dfrac{5}{2}\]
\[\Rightarrow -\dfrac{7}{2}+\dfrac{5}{2}=\dfrac{-2}{2}\]
\[\Rightarrow -1\]
Therefore, the value of the given expression is -1.
Note: The formula of the sum of the zeroes and the product of the zeroes should be correctly and carefully written without getting confused with the terms involved. We do not have to explicitly find the value of the zeroes of the polynomial \[\alpha \] and \[\beta \]. Since, the expression did not require such independent values, so used the values from the sum and the product of the zeroes and directly substituted in the expression given to us.
Complete step-by-step solution:
According to the given question, we are given the zeros of a polynomial \[2{{y}^{2}}+7y+5\] and we are asked to find the value of \[\alpha +\beta +\alpha \beta \].
The polynomial that we have is,
\[2{{y}^{2}}+7y+5\]
It is of the form, \[a{{y}^{2}}+by+c\] and so we get the corresponding values of a, b and c as,
\[a=2,b=7,c=5\]
We are given that \[\alpha ,\beta \] are the zeroes of the polynomial. So, we will first figure out the value of these zeroes and then we will find the value of the expression.
So, we have,
Sum of the zeroes = \[\alpha +\beta =-\dfrac{b}{a}\]
\[\Rightarrow \alpha +\beta =-\dfrac{7}{2}\]
Next, we have,
The product of zeroes = \[\alpha \beta =\dfrac{c}{a}\]
We get.
\[\Rightarrow \alpha \beta =\dfrac{5}{2}\]
Now, we can find the value of the given expression, that is, \[\alpha +\beta +\alpha \beta \]
So, we get,
\[\Rightarrow \left( \alpha +\beta \right)+\alpha \beta \]
We will now substitute the values in the above expression and we get,
\[\Rightarrow \left( -\dfrac{7}{2} \right)+\dfrac{5}{2}\]
\[\Rightarrow -\dfrac{7}{2}+\dfrac{5}{2}=\dfrac{-2}{2}\]
\[\Rightarrow -1\]
Therefore, the value of the given expression is -1.
Note: The formula of the sum of the zeroes and the product of the zeroes should be correctly and carefully written without getting confused with the terms involved. We do not have to explicitly find the value of the zeroes of the polynomial \[\alpha \] and \[\beta \]. Since, the expression did not require such independent values, so used the values from the sum and the product of the zeroes and directly substituted in the expression given to us.
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