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Quadratic Equation and Inequalities question

2019 · 9 Apr · Shift 1 · Q24
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  5. /2019 · 9 Apr · Shift 1 · Q24

Quadratic Equation and Inequalities question

2019 · 9 Apr · Shift 1 · Q24

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
Let p, q ∈\in∈ R. If 2 - 3\sqrt 33​ is a root of the quadratic equation, x2 + px + q = 0, then :
  1. A
    p2 – 4q – 12 = 0
  2. B
    q2 – 4p – 16 = 0
  3. C
    q2 + 4p + 14 = 0
  4. D
    p2 – 4q + 12 = 0
View written solutionFree

Correct answer: A

  1. Since the quadratic equation is x2+px+q=0x^2+px+q=0x2+px+q=0 and p,q∈Rp,q\in\mathbb Rp,q∈R, its coefficients are real.

  2. One root is given as 2−3.2-\sqrt{3}.2−3​. Because the coefficients are real, the conjugate root must also be a root: 2+3.2+\sqrt{3}.2+3​.

  3. Now form the quadratic using sum and product of roots.

Let the roots be α=2−3\alpha=2-\sqrt{3}α=2−3​ and β=2+3\beta=2+\sqrt{3}β=2+3​.

Then α+β=(2−3)+(2+3)=4,\alpha+\beta=(2-\sqrt{3})+(2+\sqrt{3})=4,α+β=(2−3​)+(2+3​)=4, αβ=(2−3)(2+3)=4−3=1.\alpha\beta=(2-\sqrt{3})(2+\sqrt{3})=4-3=1.αβ=(2−3​)(2+3​)=4−3=1.

  1. For the monic quadratic x2+px+q=0,x^2+px+q=0,x2+px+q=0, we have α+β=−p,αβ=q.\alpha+\beta=-p, \qquad \alpha\beta=q.α+β=−p,αβ=q. So, −p=4⇒p=−4,-p=4 \Rightarrow p=-4,−p=4⇒p=−4, q=1.q=1.q=1.

  2. Check each option.

Option A

p2−4q−12=(−4)2−4(1)−12=16−4−12=0.p^2-4q-12=(-4)^2-4(1)-12=16-4-12=0.p2−4q−12=(−4)2−4(1)−12=16−4−12=0. So A is true.

Option B

q2−4p−16=(1)2−4(−4)−16=1+16−16=1≠0.q^2-4p-16=(1)^2-4(-4)-16=1+16-16=1\neq 0.q2−4p−16=(1)2−4(−4)−16=1+16−16=1=0. So B is false.

Option C

q2+4p+14=(1)2+4(−4)+14=1−16+14=−1≠0.q^2+4p+14=(1)^2+4(-4)+14=1-16+14=-1\neq 0.q2+4p+14=(1)2+4(−4)+14=1−16+14=−1=0. So C is false.

Option D

p2−4q+12=16−4+12=24≠0.p^2-4q+12=16-4+12=24\neq 0.p2−4q+12=16−4+12=24=0. So D is false.

  1. Therefore, the only correct option is A.\boxed{A}.A​.
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