Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Quadratic Equation and Inequalities question

2020 · 5 Sep · Shift 1 · Q40
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Quadratic Equation and Inequalities
  5. /2020 · 5 Sep · Shift 1 · Q40

Quadratic Equation and Inequalities question

2020 · 5 Sep · Shift 1 · Q40

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
The product of the roots of the equation 9x2 - 18|x| + 5 = 0 is :
  1. A
    59{{5} \over {9}}95​
  2. B
    527{{5} \over {27}}275​
  3. C
    2581{{25} \over {81}}8125​
  4. D
    259{{25} \over {9}}925​
View written solutionFree

Correct answer: C

  1. Given equation

    9x2−18∣x∣+5=09x^2 - 18|x| + 5 = 09x2−18∣x∣+5=0

    We need the product of all roots of this equation.

  2. Substitute

    Let y=∣x∣y = |x|y=∣x∣ Then since x2=∣x∣2=y2x^2 = |x|^2 = y^2x2=∣x∣2=y2, the equation becomes 9y2−18y+5=09y^2 - 18y + 5 = 09y2−18y+5=0

  3. Solve the quadratic in yyy

    9y2−18y+5=09y^2 - 18y + 5 = 09y2−18y+5=0

    Using factorization or quadratic formula: 9y2−18y+5=(3y−1)(3y−5)=09y^2 - 18y + 5 = (3y-1)(3y-5)=09y2−18y+5=(3y−1)(3y−5)=0

    So, y=13,y=53y = \frac{1}{3}, \quad y = \frac{5}{3}y=31​,y=35​

  4. Convert back to xxx

    Since y=∣x∣y = |x|y=∣x∣,

    • If ∣x∣=13|x| = \frac{1}{3}∣x∣=31​, then x=±13x = \pm \frac{1}{3}x=±31​
    • If ∣x∣=53|x| = \frac{5}{3}∣x∣=35​, then x=±53x = \pm \frac{5}{3}x=±35​

    Therefore, the roots are 13, −13, 53, −53\frac{1}{3},\ -\frac{1}{3},\ \frac{5}{3},\ -\frac{5}{3}31​, −31​, 35​, −35​

  5. Find their product

    (13)(−13)(53)(−53)\left(\frac{1}{3}\right)\left(-\frac{1}{3}\right)\left(\frac{5}{3}\right)\left(-\frac{5}{3}\right)(31​)(−31​)(35​)(−35​)

    First pair: (13)(−13)=−19\left(\frac{1}{3}\right)\left(-\frac{1}{3}\right) = -\frac{1}{9}(31​)(−31​)=−91​

    Second pair: (53)(−53)=−259\left(\frac{5}{3}\right)\left(-\frac{5}{3}\right) = -\frac{25}{9}(35​)(−35​)=−925​

    Thus, (−19)(−259)=2581\left(-\frac{1}{9}\right)\left(-\frac{25}{9}\right)=\frac{25}{81}(−91​)(−925​)=8125​

  6. Check options

    2581\frac{25}{81}8125​ corresponds to Option C.

  7. Comparison with stored answer

    Stored correct answer: C

    Our derived answer: C

    Hence, they agree.

PreviousNext

More from Quadratic Equation and Inequalities

  • If α and β are the roots of the equation, 7x2 – 3x – 2 = 0, then the value of 1−α2α​+1−β2β​ is equal to :2020 · MCQ
  • If α and β be two roots of the equation x2 – 64x + 256 = 0. Then the value of (β5α3​)1/8+(α5β3​)1/8 is :2020 · MCQ
  • If α and β are the roots of the equation 2x(2x + 1) = 1, then β is equal to :2020 · MCQ
  • Let α and β be two real roots of the equation (k + 1)tan2x -2​. λ tanx = (1 - k), where k(e - 1) and λ are real numbers. if tan2 (α+β) = 50, then a value of λ is:2020 · MCQ
  • Let α and β be the roots of the equation x2 - x - 1 = 0. If pk =(α)k+(β)k, k ≥ 1, then which one of the following statements is not true?2020 · MCQ
  • The least positive value of 'a' for which the equation 2x2 + (a – 10)x + 233​ = 2a has real roots is2020 · Numerical
  • Let α=2−1+i3​​. If a=(1+α)k=0∑100​α2k and b=k=0∑100​α3k, then a and b are the roots of the quadratic equation :2020 · MCQ
  • Let S be the set of all real roots of the equation, 3x(3x – 1) + 2 = |3x – 1| + |3x – 2|. Then S :2020 · MCQ