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BITSAT 2022
Exam
BITSAT
Year
2022
Subject
Mathematics
Chapter
Quadratic Equation

BITSAT 2022 · Mathematics · Quadratic Equation

If α be a root of the equation 4 x^2+2 x-1=0 , then the other root...

If α\alpha be a root of the equation 4x2+2x−1=04 x^2+2 x-1=0, then the other root of the equation is

Options

  1. A4α3+2α4 \alpha^3+2 \alpha
  2. B4α2−2α4 \alpha^2-2 \alpha
  3. C4α3−3α4 \alpha^3-3 \alpha
  4. D4α3+3α4 \alpha^3+3 \alpha

Correct answer

C. 4α3−3α4 \alpha^3-3 \alpha

Step-by-step solution

4x2+2x−1=0∴α+β=−12,αβ=−144 x^2+2 x-1=0\therefore \quad \alpha+\beta=-\frac{1}{2}, \alpha \beta=-\frac{1}{4} Also, 4α2+2α−1=04 \alpha^2+2 \alpha-1=0 as α\alpha is a root and we have to prove that β=4α3−3α\beta=4 \alpha^3-3 \alpha. Now, 4α3−3α=4α2⋅α−3α=α(1−2α)−3α=−2α2−2α=−12[4α2+4α]=−12[1−2α+4α]=−12(1+2α)=−12−α=β4 \alpha^3-3 \alpha=4 \alpha^2 \cdot \alpha-3 \alpha=\alpha(1-2 \alpha)-3 \alpha=-2 \alpha^2-2 \alpha=-\frac{1}{2}\left[4 \alpha^2+4 \alpha\right]=-\frac{1}{2}[1-2 \alpha+4 \alpha]=-\frac{1}{2}(1+2 \alpha)=-\frac{1}{2}-\alpha=\beta Now, α+β=−12\alpha+\beta=-\frac{1}{2} [from Eq. (i)] Hence, β=4α3−3α\beta=4 \alpha^3-3 \alpha.
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