Question

The sum of two roots of a quadratic equation is 7 and sum of their cubes is 91,find equation

Answer 1

Answer:

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Answer 2

Answer: The required quadratic equation is $x^2 - 7x + 36 = 0$.

Step-by-step explanation:

Let the roots of the quadratic equation be $\alpha$ and $\beta$. We know that the sum of the roots is given by the expression $\alpha + \beta$. Thus, from the given information, we can write:

$$\alpha + \beta = 7$$

We also know that the sum of cubes of the roots is given by the expression $\alpha^3 + \beta^3$. Thus, we can write:

$$\alpha^3 + \beta^3 = 91$$

We can use the identity $(a^3 + b^3) = (a+b)(a^2-ab+b^2)$ to factorize the second equation as follows:

$$\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) = 7(\alpha^2 - \alpha\beta + \beta^2) = 91$$

Simplifying the above equation, we get:

$$\alpha^2 - \alpha\beta + \beta^2 = \frac{91}{7} = 13$$

We also know that the sum of the roots is given by the expression $\alpha + \beta$, and the product of the roots is given by the expression $\alpha\beta$. Thus, we can write the quadratic equation as:

$$x^2 - (\alpha + \beta)x + \alpha\beta = 0$$

Substituting the given values, we get:

$$x^2 - 7x + \alpha\beta = 0$$

Using the relation $\alpha^2 - \alpha\beta + \beta^2 = 13$, we can write:

$$\alpha\beta = (\alpha^2 + \beta^2) - (\alpha^2 - \alpha\beta + \beta^2) = (7^2 - 13) = 36$$

Substituting the value of $\alpha\beta$ in the quadratic equation, we get:

$$x^2 - 7x + 36 = 0$$

Thus, the required quadratic equation is $x^2 - 7x + 36 = 0$.

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