Verified Solution • 2025–26 Session moderate

QJEEM23Ma72d4: JEE Main 2023 Mathematics - Class 11 JEEM23Ma72d4 Solution: The sum of all the roots of the equation is :

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

Multiple Choice Options

  • A 11+311+\sqrt{3}
  • B 9+39+\sqrt{3}
  • C 9−39-\sqrt{3}
  • D 11−311-\sqrt{3}

Step-by-Step Verified Solution

amp; We have, ∣x2−8x+15∣−2x+7=0amp;⇒∣(x−3)(x−5)∣−2x+7=0\begin{aligned}& \text { We have, }\left|x^2-8 x+15\right|-2 x+7=0 \\\\& \Rightarrow |(x-3)(x-5)|-2 x+7=0\end{aligned}

JEE Main 2023 (Online) 6th April Morning Shift Mathematics - Quadratic Equation and Inequalities Question 27 English Explanation

Now, when x≤3x \leq 3 or x≥5x \geq 5 , then

amp;x2−8x+15−2x+7=0amp;⇒x2−10x+22=0amp;⇒x2−10x+25−3=0amp;⇒(x−5)2−3=0amp;⇒(x−5)=±3amp;⇒x=5+3,5−3\begin{array}{ccrl} & x^2-8 x+15-2 x+7 =0 \\\\&\Rightarrow x^2-10 x+22=0 \\\\&\Rightarrow x^2-10 x+25-3=0 \\\\&\Rightarrow (x-5)^2-3=0 \\\\&\Rightarrow (x-5)= \pm \sqrt{3} \\\\&\Rightarrow x=5+\sqrt{3}, 5-\sqrt{3}\end{array}

Since, x≤3x \leq 3 or x≥5x \geq 5

∴x=5+3\therefore x=5+\sqrt{3}

When, 3 < x < 5 , then

amp;−(x2−8x+15)−2x+7=0amp;⇒−x2+8x−15−2x+7=0amp;⇒−x2+6x−8=0amp;⇒x2−6x+8=0amp;⇒(x−4)(x−2)=0⇒x=2,4\begin{array}{rlrl} & -\left(x^2-8 x+15\right)-2 x+7 =0 \\\\& \Rightarrow -x^2+8 x-15-2 x+7 =0 \\\\& \Rightarrow -x^2+6 x-8 =0 \\\\& \Rightarrow x^2-6 x+8 =0 \\\\& \Rightarrow (x-4)(x-2) =0 \Rightarrow x=2,4\end{array}

Since, 3 < x < 5

∴x=4\therefore x=4

∴ Sum of roots =(5+3)+4=9+3\therefore \text { Sum of roots }=(5+\sqrt{3})+4=9+\sqrt{3}

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