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Quadratic Equation and Inequalities question

2019 · 8 Apr · Shift 1 · Q25
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  5. /2019 · 8 Apr · Shift 1 · Q25

Quadratic Equation and Inequalities question

2019 · 8 Apr · Shift 1 · Q25

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
The sum of the solutions of the equation ∣x−2∣+x(x−4)+2=0\left| {\sqrt x - 2} \right| + \sqrt x \left( {\sqrt x - 4} \right) + 2 = 0​x​−2​+x​(x​−4)+2=0 (x > 0) is equal to:
  1. A
    9
  2. B
    12
  3. C
    4
  4. D
    10
View written solutionFree

Correct answer: D

  1. Substitute

Let t=x,t>0t = \sqrt{x}, \quad t>0t=x​,t>0 Then the equation becomes ∣t−2∣+t(t−4)+2=0|t-2| + t(t-4) + 2 = 0∣t−2∣+t(t−4)+2=0 that is, ∣t−2∣+t2−4t+2=0.|t-2| + t^2 - 4t + 2 = 0.∣t−2∣+t2−4t+2=0.

  1. Case 1: t≥2t \ge 2t≥2

Then ∣t−2∣=t−2.|t-2| = t-2.∣t−2∣=t−2. So the equation becomes t−2+t2−4t+2=0t-2 + t^2 - 4t + 2 = 0t−2+t2−4t+2=0 t2−3t=0t^2 - 3t = 0t2−3t=0 t(t−3)=0.t(t-3)=0.t(t−3)=0. Since t>0t>0t>0 and t≥2t\ge 2t≥2, we get t=3.t=3.t=3. Thus x=t2=9.x=t^2=9.x=t2=9.

  1. Case 2: 0<t<20<t<20<t<2

Then ∣t−2∣=2−t.|t-2| = 2-t.∣t−2∣=2−t. So the equation becomes 2−t+t2−4t+2=02-t + t^2 - 4t + 2 = 02−t+t2−4t+2=0 t2−5t+4=0t^2 - 5t + 4 = 0t2−5t+4=0 (t−1)(t−4)=0.(t-1)(t-4)=0.(t−1)(t−4)=0. In the interval 0<t<20<t<20<t<2, only t=1t=1t=1 is valid. Thus x=t2=1.x=t^2=1.x=t2=1.

  1. Sum of solutions

The solutions are x=1,  9.x=1,\;9.x=1,9. Hence their sum is 1+9=10.1+9=10.1+9=10.

  1. Compare with stored answer

Derived answer: 10

Option D is correct, which matches the stored correct answer.

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