JEE MainMathematicsQuadratic Equation and InequalitiesNumerical+4 / −1
The number of real roots of the equation e4x e3x 4e2x ex + 1 = 0 is equal to .
Numerical answer
View written solutionFree
Correct answer: 2
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Interpret the equation
The given equation is
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Substitute to reduce it to a polynomial
Let Since for all real , we must have
Then the equation becomes
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Factor the quartic
We try factoring it into two quadratics:
Expanding,
= t^4 + (a+b)t^3 + (ab-2)t^2 - (a+b)t + 1.$$ Comparing coefficients with $$t^4 - t^3 - 4t^2 - t + 1,$$ we get $$a+b = -1, ab - 2 = -4 \,\Rightarrow\, ab = -2.$$ Numbers satisfying these are $a=1$, $b=-2$ (or vice versa). Hence $$t^4 - t^3 - 4t^2 - t + 1 = (t^2+t-1)(t^2-2t-1).$$ -
Solve each quadratic
From we get Only the positive root is valid:
From we get Only the positive root is valid:
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Convert back to
Since , each positive value of gives exactly one real value of :
Therefore, the equation has 2 real roots.
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Comparison with stored answer
My derived answer is , which matches the stored correct answer.
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