
If $$\mathrm\alpha\;\mathrm{and}\;\mathrm\beta$$ are the zeroes of the quadratic polynomial f(x) = $x^2+ bx + c$, then evaluate the following-
$$\mathrm\alpha^2\mathrm\beta+\mathrm{\mathrm\alpha\mathrm\beta}^2$$
Answer
616.2k+ views
Hint: This is a question of quadratic equations, we have to use the formulas-
$$\mathrm\alpha+\mathrm\beta=-\dfrac{\mathrm b}{\mathrm a}\\\mathrm{\mathrm\alpha\mathrm\beta}=\dfrac{\mathrm c}{\mathrm a}$$
Evaluate the given function in such a way that we can substitute the above given formula to get the solution.
Complete step-by-step answer:
We have to convert the given expression such that it can be expressed in the form of $$\mathrm\alpha+\mathrm\beta\;\mathrm{and}\;\mathrm{\mathrm\alpha\mathrm\beta}$$ only, so that we can apply the given formulas, substitute the values and find the result.
So we will convert the equation as follows-
$$\mathrm{Taking}\;\mathrm{\mathrm\alpha\mathrm\beta}\;\mathrm{common},\;\\=\mathrm{\mathrm\alpha\mathrm\beta}\left(\mathrm\alpha+\mathrm\beta\right)$$
Now we will substitute the given formula in this expression to find its value using-
$$\mathrm\alpha+\mathrm\beta=-\dfrac{\mathrm b}{\mathrm a}\\\mathrm{\mathrm\alpha\mathrm\beta}=\dfrac{\mathrm c}{\mathrm a}$$
$$=\dfrac{\mathrm c}{\mathrm a}\left(-\dfrac{\mathrm b}{\mathrm a}\right)\\=-\dfrac{\mathrm{bc}}{\mathrm a^2}$$
This is the required answer.
Note: The above given formula signifies that the sum of roots of a quadratic equation is the negative of the ratio of coefficient of x and $x^2$. Also, the product of roots is the ratio of constant term and coefficient of $x^2$. It is also recommended to simplify the final answer.
$$\mathrm\alpha+\mathrm\beta=-\dfrac{\mathrm b}{\mathrm a}\\\mathrm{\mathrm\alpha\mathrm\beta}=\dfrac{\mathrm c}{\mathrm a}$$
Evaluate the given function in such a way that we can substitute the above given formula to get the solution.
Complete step-by-step answer:
We have to convert the given expression such that it can be expressed in the form of $$\mathrm\alpha+\mathrm\beta\;\mathrm{and}\;\mathrm{\mathrm\alpha\mathrm\beta}$$ only, so that we can apply the given formulas, substitute the values and find the result.
So we will convert the equation as follows-
$$\mathrm{Taking}\;\mathrm{\mathrm\alpha\mathrm\beta}\;\mathrm{common},\;\\=\mathrm{\mathrm\alpha\mathrm\beta}\left(\mathrm\alpha+\mathrm\beta\right)$$
Now we will substitute the given formula in this expression to find its value using-
$$\mathrm\alpha+\mathrm\beta=-\dfrac{\mathrm b}{\mathrm a}\\\mathrm{\mathrm\alpha\mathrm\beta}=\dfrac{\mathrm c}{\mathrm a}$$
$$=\dfrac{\mathrm c}{\mathrm a}\left(-\dfrac{\mathrm b}{\mathrm a}\right)\\=-\dfrac{\mathrm{bc}}{\mathrm a^2}$$
This is the required answer.
Note: The above given formula signifies that the sum of roots of a quadratic equation is the negative of the ratio of coefficient of x and $x^2$. Also, the product of roots is the ratio of constant term and coefficient of $x^2$. It is also recommended to simplify the final answer.
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