Question

From the quadratic equation whose roots $\alpha$ and $\beta$ satisfy the relations $\alpha \beta = 768$ and ${\alpha ^2} + {\beta ^2} = 1600$?

Answer 1

To do this question, we should know how to write the quadratic equation in terms of sum of roots and product of roots. Also, here we have to use the formula ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$to find the value of the sum of roots.

Complete step by step answer:
In the above question, it is given that $\alpha$ and $\beta$ are the roots of the equation.
If $\alpha$ and $\beta$ are the roots of a quadratic equation, the equation can be written as:
${x^2} - \left( {\alpha + \beta } \right)x + \alpha \beta = 0$
We are given
$\alpha \beta = 768$ and ${\alpha ^2} + {\beta ^2} = 1600$
Now,
${\left( {\alpha + \beta } \right)^2} = {\alpha ^2} + {\beta ^2} + 2\alpha \beta$
Now, substituting the values of $\alpha \beta$ and ${\alpha ^2} + {\beta ^2}$ in the above equation.
${\left( {\alpha + \beta } \right)^2} = 1600 + 2\left( {768} \right)$
${\left( {\alpha + \beta } \right)^2} = 1600 + 2\left( {768} \right) = 3136$
Now taking square root both sides
$\alpha + \beta = \sqrt {3136}$
$\alpha + \beta = 56$
Now, substitute the values of $\alpha + \beta$ and $\alpha \beta$ in the quadratic equation.
We get,
${x^2} - 56x + 768 = 0$
Therefore, our required quadratic equation is ${x^2} - 56x + 768 = 0$.

Note: Thus, the sum of roots of a quadratic equation is given by the negative ratio of coefficient of x and ${x^2}$. The product of roots is given by the ratio of the constant term and the coefficient of ${x^2}$. We know that the graph of a quadratic function is represented using a parabola. If α and β are the real roots of a quadratic equation, then the point of intersection of the plot of this function with the x-axis represents its roots.