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Quadratic Equation and Inequalities question

2005 · Shift 0 · Q68
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  5. /2005 · Shift 0 · Q68

Quadratic Equation and Inequalities question

2005 · Shift 0 · Q68

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
If the roots of the equation x2−bx+c=0{x^2} - bx + c = 0x2−bx+c=0 be two consecutive integers, then b2−4c{b^2} - 4cb2−4c equals
  1. A
    −2-2−2
  2. B
    333
  3. C
    222
  4. D
    111
View written solutionFree

Correct answer: D

  1. Let the two consecutive integer roots be nnn and n+1n+1n+1, where nnn is an integer.

  2. For the quadratic equation x2−bx+c=0,x^2-bx+c=0,x2−bx+c=0, by Vieta’s formulas:

    • Sum of roots =b= b=b
    • Product of roots =c= c=c
  3. Using the roots nnn and n+1n+1n+1: b=n+(n+1)=2n+1b=n+(n+1)=2n+1b=n+(n+1)=2n+1 c=n(n+1)=n2+nc=n(n+1)=n^2+nc=n(n+1)=n2+n

  4. Now compute: b2−4c=(2n+1)2−4(n2+n)b^2-4c=(2n+1)^2-4(n^2+n)b2−4c=(2n+1)2−4(n2+n)

  5. Expanding: b2−4c=4n2+4n+1−4n2−4n=1b^2-4c=4n^2+4n+1-4n^2-4n=1b2−4c=4n2+4n+1−4n2−4n=1

  6. Therefore, b2−4c=1b^2-4c=1b2−4c=1

  7. Checking options:

    • A: −2-2−2 ❌
    • B: 333 ❌
    • C: 222 ❌
    • D: 111 ✅

Hence, the correct answer is D.

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