JEE MAIN - Mathematics (2023 - 6th April Morning Shift - No. 11)
The sum of all the roots of the equation $$\left|x^{2}-8 x+15\right|-2 x+7=0$$ is :
$$11+\sqrt{3}$$
$$9+\sqrt{3}$$
$$9-\sqrt{3}$$
$$11-\sqrt{3}$$
Explanation
$$
\begin{aligned}
& \text { We have, }\left|x^2-8 x+15\right|-2 x+7=0 \\\\
& \Rightarrow |(x-3)(x-5)|-2 x+7=0
\end{aligned}
$$
_6th_April_Morning_Shift_en_11_1.png)
Now, when $x \leq 3$ or $x \geq 5$, then
$$ \begin{array}{ccrl} & x^2-8 x+15-2 x+7 =0 \\\\ &\Rightarrow x^2-10 x+22=0 \\\\ &\Rightarrow x^2-10 x+25-3=0 \\\\ &\Rightarrow (x-5)^2-3=0 \\\\ &\Rightarrow (x-5)= \pm \sqrt{3} \\\\ &\Rightarrow x=5+\sqrt{3}, 5-\sqrt{3} \end{array} $$
Since, $x \leq 3$ or $x \geq 5$
$$ \therefore x=5+\sqrt{3} $$
When, $3 < x < 5$, then
$$ \begin{array}{rlrl} & -\left(x^2-8 x+15\right)-2 x+7 =0 \\\\ & \Rightarrow -x^2+8 x-15-2 x+7 =0 \\\\ & \Rightarrow -x^2+6 x-8 =0 \\\\ & \Rightarrow x^2-6 x+8 =0 \\\\ & \Rightarrow (x-4)(x-2) =0 \Rightarrow x=2,4 \end{array} $$
Since, $3 < x < 5$
$$ \therefore x=4 $$
$$ \therefore \text { Sum of roots }=(5+\sqrt{3})+4=9+\sqrt{3} $$
Now, when $x \leq 3$ or $x \geq 5$, then
$$ \begin{array}{ccrl} & x^2-8 x+15-2 x+7 =0 \\\\ &\Rightarrow x^2-10 x+22=0 \\\\ &\Rightarrow x^2-10 x+25-3=0 \\\\ &\Rightarrow (x-5)^2-3=0 \\\\ &\Rightarrow (x-5)= \pm \sqrt{3} \\\\ &\Rightarrow x=5+\sqrt{3}, 5-\sqrt{3} \end{array} $$
Since, $x \leq 3$ or $x \geq 5$
$$ \therefore x=5+\sqrt{3} $$
When, $3 < x < 5$, then
$$ \begin{array}{rlrl} & -\left(x^2-8 x+15\right)-2 x+7 =0 \\\\ & \Rightarrow -x^2+8 x-15-2 x+7 =0 \\\\ & \Rightarrow -x^2+6 x-8 =0 \\\\ & \Rightarrow x^2-6 x+8 =0 \\\\ & \Rightarrow (x-4)(x-2) =0 \Rightarrow x=2,4 \end{array} $$
Since, $3 < x < 5$
$$ \therefore x=4 $$
$$ \therefore \text { Sum of roots }=(5+\sqrt{3})+4=9+\sqrt{3} $$
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