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JEE Main 20236 AprilMorning Shift
Exam
JEE Main
Year
2023
Date
6 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Quadratic Equation

JEE Main 2023 · 6 April · Morning Shift · Mathematics · Quadratic Equation

The sum of all the roots of the equation | x ^ 2 - 8x + 15| - 2x +...

The sum of all the roots of the equation ∣x2−8x+15∣−2x+7=0\left|{x}^{2} - 8x + 15\right| - 2x + 7 = 0 is

Options

  1. A9−39 - \sqrt{3}
  2. B9+39 + \sqrt{3}
  3. C11−311 - \sqrt{3}
  4. D11+311 + \sqrt{3}

Correct answer

B. 9+39 + \sqrt{3}

Step-by-step solution

Given,

∣x2−8x+15∣−2x+7=0\left|{x}^{2} - 8x + 15\right| - 2x + 7 = 0

Now taking Case:II when,

x2−8x+15≥0⇒(x−3)(x−5)≥0{x}^{2} - 8x + 15 \ge 0 ⇒ ( x - 3 ) ( x - 5 ) \ge 0

⇒x≤3⇒ x \le 3 OR x≥5x \ge 5

\left|{x}^{2} - 8x + 15\right| - 2x + 7 = 0</p><p></p><p>⇒ {x}^{2} - 8x + 15 - 2x + 7 = 0</p><p></p><p>⇒ {x}^{2} - 10x + 22 = 0</p><p></p><p>⇒ x = 5 + \sqrt{3}, 5 - \sqrt{3}\left(\text{rejected}\right)</p><p>So,</p><p>So,\alpha = 5 + \sqrt{3}</p><p>NowtakingCase:</p><p>Now taking Case:IIwhen,</p><p>∣x2−8x+15∣−2x+7=0when,</p><p>\left|{x}^{2} - 8x + 15\right| - 2x + 7 = 0

⇒x2−8x+15<0⇒3<x<5⇒ {x}^{2} - 8x + 15 < 0 ⇒ 3 < x < 5

⇒x2−8x+15+2x−7=0⇒ {x}^{2} - 8x + 15 + 2x - 7 = 0

⇒x2−6x+8=0⇒x=4,2(rejected)⇒ {x}^{2} - 6x + 8 = 0 ⇒ x = 4, 2\left(\text{rejected}\right)

So, γ=4\gamma = 4

Hence, α+γ=9+3\alpha + \gamma = 9 + \sqrt{3}

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