Question:
If the equation $ (a+1){{x}^{2}}-(a+2)x+(a+3)=0 $ has roots equal in magnitude but opposite in signs, then the roots of the equation are
If the equation $ (a+1){{x}^{2}}-(a+2)x+(a+3)=0 $ has roots equal in magnitude but opposite in signs, then the roots of the equation are
Updated On: Jun 3, 2024
- $ \pm \,\,a $
- $ \pm \,\frac{1}{2}\,a $
- $ \pm \,\frac{3}{2}\,a $
- $ \pm \,2\,a $
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The Correct Option is B
Solution and Explanation
Given equation is $ (a+1){{x}^{2}}-(a+2)x+(a+3)=0 $
Since, roots are equal in magnitude and opposite in sign.
$ \therefore $ coefficient of x is zero ie, $ a+2=0 $
$ \Rightarrow $ $ a=-2 $ ..(i)
$ \therefore $ Equation is $ (-2+1){{x}^{2}}-(-2+2)x+(-2+3)=0 $
$ \Rightarrow $ $ -{{x}^{2}}+1=0 $
$ \Rightarrow $ $ x=\pm 1 $ ..(ii)
Only option [b] ie, $ \pm \frac{1}{2} $
a satisfies Eqs. (i) and (ii).
Since, roots are equal in magnitude and opposite in sign.
$ \therefore $ coefficient of x is zero ie, $ a+2=0 $
$ \Rightarrow $ $ a=-2 $ ..(i)
$ \therefore $ Equation is $ (-2+1){{x}^{2}}-(-2+2)x+(-2+3)=0 $
$ \Rightarrow $ $ -{{x}^{2}}+1=0 $
$ \Rightarrow $ $ x=\pm 1 $ ..(ii)
Only option [b] ie, $ \pm \frac{1}{2} $
a satisfies Eqs. (i) and (ii).
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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.
