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Quadratic Equation and Inequalities question

2025 · 24 Jan · Shift 1 · Q28
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Quadratic Equation and Inequalities question

2025 · 24 Jan · Shift 1 · Q28

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
The product of all the rational roots of the equation (x2−9x+11)2−(x−4)(x−5)=3\left(x^2-9 x+11\right)^2-(x-4)(x-5)=3(x2−9x+11)2−(x−4)(x−5)=3, is equal to
  1. A
    7
  2. B
    21
  3. C
    28
  4. D
    14
View written solutionFree

Correct answer: D

  1. Let y=x2−9x+11.y=x^2-9x+11.y=x2−9x+11. Then the equation becomes y2−(x−4)(x−5)=3.y^2-(x-4)(x-5)=3.y2−(x−4)(x−5)=3. Since (x−4)(x−5)=x2−9x+20=y+9,(x-4)(x-5)=x^2-9x+20 = y+9,(x−4)(x−5)=x2−9x+20=y+9, because y=x2−9x+11,y=x^2-9x+11,y=x2−9x+11, so x2−9x+20=(x2−9x+11)+9=y+9.x^2-9x+20=(x^2-9x+11)+9=y+9.x2−9x+20=(x2−9x+11)+9=y+9.

  2. Substitute this into the equation: y2−(y+9)=3.y^2-(y+9)=3.y2−(y+9)=3. So, y2−y−12=0.y^2-y-12=0.y2−y−12=0. Factorizing, (y−4)(y+3)=0.(y-4)(y+3)=0.(y−4)(y+3)=0. Hence, y=4ory=−3.y=4 \quad \text{or} \quad y=-3.y=4ory=−3.

  3. Now solve each quadratic in xxx.

For y=4y=4y=4: x2−9x+11=4x^2-9x+11=4x2−9x+11=4 x2−9x+7=0.x^2-9x+7=0.x2−9x+7=0. Its discriminant is Δ=81−28=53,\Delta=81-28=53,Δ=81−28=53, which is not a perfect square, so its roots are irrational.

For y=−3y=-3y=−3: x2−9x+11=−3x^2-9x+11=-3x2−9x+11=−3 x2−9x+14=0.x^2-9x+14=0.x2−9x+14=0. Factorizing, (x−7)(x−2)=0.(x-7)(x-2)=0.(x−7)(x−2)=0. So the rational roots are x=7,  2.x=7,\;2.x=7,2.

  1. Product of all rational roots: 7⋅2=14.7\cdot 2=14.7⋅2=14.

Therefore, the correct option is 14.\boxed{14}.14​.

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