Number of real roots of 2 x 99 + 3 x 98 + 2 x 97 + 3 x 96 + …… + 2 x + 3 = 0 is

Given equation can be written as ( 2 x + 3 ) ( x 98 + x 96 + …… + 1 ) = ( 2 x + 3 ) x 2 − 1 ( x 100 − 1 ) So, the real roots are x = ± 1 , 2 − 3 , out of which ± 1 are not roots of given equation.

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MathematicsQuadratic EquationTheory of equationsEasy2 minQB

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Number of real roots of $2 x^{99} + 3 x^{98} + 2 x^{97} + 3 x^{96} + \dots\dots + 2 x + 3 = 0$ is

Answer:

Solution:

Given equation can be written as $(2 x + 3) (x^{98} + x^{96} + \dots\dots + 1) = (2 x + 3) \frac{( x ^{100} - 1 )}{x ^{2} - 1}$

So, the real roots are $x = \pm 1, \frac{- 3}{2}$ , out of which $\pm 1$ are not roots of given equation.

Stream:JEESubject:MathematicsTopic:Quadratic EquationSubtopic:Theory of equations

⏱ 2mℹ️ Source: QB

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