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Quadratic Equation and Inequalities question

2004 · Shift 0 · Q101
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Quadratic Equation and Inequalities question

2004 · Shift 0 · Q101

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
If (1−p)\left( {1 - p} \right)(1−p) is a root of quadratic equation x2+px+(1−p)=0{x^2} + px + \left( {1 - p} \right) = 0x2+px+(1−p)=0 then its root are
  1. A
    −1,2- 1,2−1,2
  2. B
    −1,1- 1,1−1,1
  3. C
    0,−10,-10,−1
  4. D
    0,10,10,1
View written solutionFree

Correct answer: C

  1. Given quadratic equation

    x2+px+(1−p)=0x^2 + px + (1-p)=0x2+px+(1−p)=0

    It is given that (1−p)(1-p)(1−p) is one root.

  2. Substitute the root into the equation

    If x=1−px=1-px=1−p is a root, then

    (1−p)2+p(1−p)+(1−p)=0(1-p)^2 + p(1-p) + (1-p)=0(1−p)2+p(1−p)+(1−p)=0

  3. Simplify

    First expand:

    (1−p)2=1−2p+p2(1-p)^2 = 1-2p+p^2(1−p)2=1−2p+p2 p(1−p)=p−p2p(1-p)=p-p^2p(1−p)=p−p2

    So,

    1−2p+p2+p−p2+1−p=01-2p+p^2 + p-p^2 + 1-p = 01−2p+p2+p−p2+1−p=0

    Combine like terms:

    2−2p=02 - 2p = 02−2p=0

    2(1−p)=02(1-p)=02(1−p)=0

    p=1p=1p=1

  4. Form the quadratic with p=1p=1p=1

    Substitute p=1p=1p=1 into the equation:

    x2+x+(1−1)=0x^2 + x + (1-1)=0x2+x+(1−1)=0

    x2+x=0x^2 + x = 0x2+x=0

    Factorize:

    x(x+1)=0x(x+1)=0x(x+1)=0

  5. Find the roots

    x=0, −1x=0, \, -1x=0,−1

  6. Match with options

    The roots are 000 and −1-1−1.

    So the correct option is:

    C: 0,−10,-10,−1

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