Q.47 The positive root of the equation x³ + x² − 2 = 0 is ________.
The positive root of the equation x3+x2−2=0 is exactly 1.
Solving the Equation
Apply the Rational Root Theorem, which states possible rational roots are factors of the constant term (-2) divided by factors of the leading coefficient (1), so ±1, ±2. Test these values in f(x)=x3+x2−2.
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f(1)=13+12−2=1+1−2=0, so x=1 is a root.
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f(−1)=(−1)3+(−1)2−2=−1+1−2=−2≠0
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f(2)=23+22−2=8+4−2=10≠0
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f(−2)=(−2)3+(−2)2−2=−8+4−2=−6≠0
Since x=1 works, factor as (x−1)(x2+2x+2)=0 using synthetic division.
Factoring Completely
Synthetic division with root 1:
1 | 1 1 0 -2
| 1 2 2
—————
1 2 2 0
Quotient is x2+2x+2=0, discriminant 4−8=−4<0, so complex roots −1±i. Thus roots are x=1, x=−1−i, x=−1+i. Only positive root is 1.
The positive root of the equation x³ + x² – 2 = 0 is a common cubic equation problem in competitive exams like CSIR NET Life Sciences (Mathematics section) and engineering mathematics. This positive root of x³ + x² – 2 = 0 equals exactly 1, verified through rational root testing and polynomial factorization.
Why Focus on the Positive Root?
Cubic equations like x³ + x² – 2 = 0 always have at least one real root. Here, sign changes in Descartes’ Rule of Signs (f(0) = -2 < 0, f(1) = 0) confirm one positive real root. Complex roots don’t qualify as positive.
Step-by-Step Solution Method
Follow these steps to find the positive root of x³ + x² – 2 = 0:
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Step 1: List possible rational roots: ±1, ±2.
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Step 2: Evaluate f(1) = 0, confirming x = 1 as root.
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Step 3: Synthetic division yields (x – 1)(x² + 2x + 2) = 0.
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Step 4: Quadratic has no real positive roots (discriminant negative).
For numerical verification, iteration methods converge to ≈1.000 from initial guess x₀ = 1.
Exam Tips for CSIR NET
Practice similar cubics: test rational roots first to save time. Verify with graphing or numerical solvers if needed.
