Question

The sum of roots of |x² - 8x + 15| - 2x + 7 = 0 is:

• A. \(11+\sqrt{3}\)
• B. \(11-\sqrt{3}\)
• C. \(9+\sqrt{3}\)
• D. \(9-\sqrt{3}\)

Answer 1

The Correct Option is C

\(|x^{2}-8x+15|=2x-7\)

Case-I : \(x\geq 5\)

\(x^{2}-10x+22=0\)

\(x=\frac{10\pm \sqrt{12}}{2}=5\pm \sqrt{3}\)

then \(x= 5+\sqrt{3}\)

Case-II : \(\frac{7}{2}\leq x\leq 5\)

\(x^{2}-8x+15=7-2x\)

\(x^{2}-6x+8=0\)

\(x=4\)

Therefore, the sum of the roots \(= 5+\sqrt{3}+4=9+\sqrt{3}\)
So, The correct option is (C): 9 + \(\sqrt{3}\)

Answer 2

Solving an Absolute Value Equation 

We have the equation $|x^2 - 8x + 15| - 2x + 7 = 0$. We can rewrite $x^2 - 8x + 15$ as $(x - 3)(x - 5)$.

Case 1: $x^2 - 8x + 15 \geq 0$. This occurs when $x \leq 3$ or $x \geq 5$.

$x^2 - 8x + 15 - 2x + 7 = 0 \Rightarrow x^2 - 10x + 22 = 0$.

The roots are given by:
$x = \frac{10 \pm \sqrt{100 - 4(22)}}{2} = \frac{10 \pm \sqrt{12}}{2} = 5 \pm \sqrt{3}$.

Since $5 + \sqrt{3} > 5$ and $5 - \sqrt{3} \approx 3.268$, the only acceptable root in this case is $5 + \sqrt{3}$.

Case 2: $x^2 - 8x + 15 < 0$. This occurs when $3 < x < 5$.

$-(x^2 - 8x + 15) - 2x + 7 = 0 \Rightarrow -x^2 + 8x - 15 - 2x + 7 = 0$.

$-x^2 + 6x - 8 = 0 \Rightarrow x^2 - 6x + 8 = 0 \Rightarrow (x - 2)(x - 4) = 0$.

The roots are x = 2 and x = 4.
Since $3 < x < 5$, only x = 4 is a valid root.

Therefore, the roots of the given equation are $5 + \sqrt{3}$ and 4.

The sum of the roots is $4 + 5 + \sqrt{3} = 9 + \sqrt{3}$.