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Quadratic Equation and Inequalities question

2004 · Shift 0 · Q100
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  5. /2004 · Shift 0 · Q100

Quadratic Equation and Inequalities question

2004 · Shift 0 · Q100

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
Let two numbers have arithmetic mean 9 and geometric mean 4. Then these numbers are the roots of the quadratic equation
  1. A
    x2−18x−16=0{x^2} - 18x - 16 = 0x2−18x−16=0
  2. B
    x2−18x+16=0{x^2} - 18x + 16 = 0x2−18x+16=0
  3. C
    x2+18x−16=0{x^2} + 18x - 16 = 0x2+18x−16=0
  4. D
    x2+18x+16=0{x^2} + 18x + 16 = 0x2+18x+16=0
View written solutionFree

Correct answer: B

  1. Let the two numbers be α\alphaα and β\betaβ.

  2. Their arithmetic mean is 999: α+β2=9\frac{\alpha+\beta}{2}=92α+β​=9 Hence, α+β=18\alpha+\beta=18α+β=18

  3. Their geometric mean is 444: αβ=4\sqrt{\alpha\beta}=4αβ​=4 Squaring both sides, αβ=16\alpha\beta=16αβ=16

  4. If two numbers are roots of a quadratic equation, then the monic quadratic is x2−(sum of roots)x+(product of roots)=0x^2-(\text{sum of roots})x+(\text{product of roots})=0x2−(sum of roots)x+(product of roots)=0

    Substituting α+β=18\alpha+\beta=18α+β=18 and αβ=16\alpha\beta=16αβ=16, x2−18x+16=0x^2-18x+16=0x2−18x+16=0

  5. Compare with the options:

    • A: x2−18x−16=0x^2-18x-16=0x2−18x−16=0
    • B: x2−18x+16=0x^2-18x+16=0x2−18x+16=0
    • C: x2+18x−16=0x^2+18x-16=0x2+18x−16=0
    • D: x2+18x+16=0x^2+18x+16=0x2+18x+16=0

    Therefore, the correct option is B.

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