Question:
If $\alpha$ and $\beta$ are $(\alpha < \beta)$ the roots of the equation $x^2+bx+c=0,$ where $c < 0 < b$,the
If $\alpha$ and $\beta$ are $(\alpha < \beta)$ the roots of the equation $x^2+bx+c=0,$ where $c < 0 < b$,the
Updated On: Aug 25, 2023
0
\(\alpha<0\)
\(\alpha\)
\(\alpha<0\)
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The Correct Option is B
Solution and Explanation
The correct answer is B:\(\alpha<0\)
Equation is \(x^2+bx++c=0,\)\(c<0<b\)
roots of this quadratic equation are \(\alpha,\beta(\alpha<\beta)\)
So,we have,
\(\alpha+\beta=-b,\,\,\,\,,\,\,\,\,-\alpha\beta=c\) here, \(c<0,\alpha\beta<0\)
For this case
\(b>0\space (given)\)
\(\therefore \)\(\alpha+\beta<0\)
\(\alpha=positive,\beta=negative,|\beta|<|\alpha|\)
\(\therefore\)\(\alpha<0\space and \space 0<\beta\)
\(\therefore\)\(\alpha<0<\beta<|\alpha|\)
Equation is \(x^2+bx++c=0,\)\(c<0<b\)
roots of this quadratic equation are \(\alpha,\beta(\alpha<\beta)\)
So,we have,
\(\alpha+\beta=-b,\,\,\,\,,\,\,\,\,-\alpha\beta=c\) here, \(c<0,\alpha\beta<0\)
For this case
\(b>0\space (given)\)
\(\therefore \)\(\alpha+\beta<0\)
\(\alpha=positive,\beta=negative,|\beta|<|\alpha|\)
\(\therefore\)\(\alpha<0\space and \space 0<\beta\)
\(\therefore\)\(\alpha<0<\beta<|\alpha|\)
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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.
