Question:
If $\alpha, \beta $ be the roots of $x^{2}-a(x-1)+b=0$, then the value of $\frac{1}{\alpha^{2}-a\alpha}+\frac{1}{\beta^{2}-a\beta}+\frac{2}{a+b}$ is
If $\alpha, \beta $ be the roots of $x^{2}-a(x-1)+b=0$, then the value of $\frac{1}{\alpha^{2}-a\alpha}+\frac{1}{\beta^{2}-a\beta}+\frac{2}{a+b}$ is
Updated On: Apr 26, 2024
- $\frac{4}{a+b}$
- $\frac{1}{a+b}$
- $0$
- $-1$
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The Correct Option is C
Solution and Explanation
Since, $\alpha$ and $\beta$ are the roots of $x^{2} -ax + a+ b = 0$, then
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Concepts Used:
Complex Numbers and Quadratic Equations
Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.
Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.
