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Quadratic Equation and Inequalities question

2020 · 6 Sep · Shift 2 · Q32
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  5. /2020 · 6 Sep · Shift 2 · Q32

Quadratic Equation and Inequalities question

2020 · 6 Sep · Shift 2 · Q32

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
If α\alphaα and β\betaβ are the roots of the equation 2x(2x + 1) = 1, then β\betaβ is equal to :
  1. A
    −2α(α+1)- 2\alpha \left( {\alpha + 1} \right)−2α(α+1)
  2. B
    2α(α+1)2\alpha \left( {\alpha + 1} \right)2α(α+1)
  3. C
    2α22{\alpha ^2}2α2
  4. D
    2α(α−1)2\alpha \left( {\alpha - 1} \right)2α(α−1)
View written solutionFree

Correct answer: A

  1. Rewrite the given equation

The equation is 2x(2x+1)=12x(2x+1)=12x(2x+1)=1

Expanding: 4x2+2x=14x^2+2x=14x2+2x=1 4x2+2x−1=04x^2+2x-1=04x2+2x−1=0

So α\alphaα and β\betaβ are the roots of 4x2+2x−1=04x^2+2x-1=04x2+2x−1=0

  1. Use the fact that α\alphaα is a root

Since α\alphaα satisfies the equation, 4α2+2α−1=04\alpha^2+2\alpha-1=04α2+2α−1=0

Hence, 4α2+2α=14\alpha^2+2\alpha=14α2+2α=1

Now factor the left side: 2α(2α+1)=12\alpha(2\alpha+1)=12α(2α+1)=1

  1. Use sum of roots

For the quadratic 4x2+2x−1=04x^2+2x-1=04x2+2x−1=0 we have α+β=−24=−12\alpha+\beta=-\frac{2}{4}=-\frac12α+β=−42​=−21​

Thus, β=−12−α\beta=-\frac12-\alphaβ=−21​−α

  1. Transform option A

Option A is −2α(α+1)=−2α2−2α-2\alpha(\alpha+1)=-2\alpha^2-2\alpha−2α(α+1)=−2α2−2α

From 4α2+2α−1=04\alpha^2+2\alpha-1=04α2+2α−1=0 we get 2α2+α=122\alpha^2+\alpha=\frac122α2+α=21​

So, −2α2−2α=−(2α2+α)−α=−12−α-2\alpha^2-2\alpha=-(2\alpha^2+\alpha)-\alpha=-\frac12-\alpha−2α2−2α=−(2α2+α)−α=−21​−α

But from step 3, β=−12−α\beta=-\frac12-\alphaβ=−21​−α

Therefore, β=−2α(α+1)\beta=-2\alpha(\alpha+1)β=−2α(α+1)

  1. Conclusion

The correct option is: A\boxed{A}A​

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