JEE Mathematics 2023 Question with Solution

Given: $|x^2 - 8x + 15| - 2x + 7 = 0$

Find: The sum of all roots.

Rewrite the equation as

$$|x^2 - 8x + 15| = 2x - 7$$

Now solve by cases based on the sign of $x^2 - 8x + 15$.

Case I: $x \geq 5$

Then

$$x^2 - 8x + 15 = 2x - 7$$

So,

$$x^2 - 10x + 22 = 0$$

Hence,

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

Among these, only

$$x = 5 + \sqrt{3}$$

is valid in this case.

Case II: $\frac{7}{2} \leq x \leq 5$

Then

$$x^2 - 8x + 15 = 7 - 2x$$

So,

$$x^2 - 6x + 8 = 0$$

Thus,

$$x = 4$$

which is valid in this interval.

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

Their sum is

$$4 + 5 + \sqrt{3} = 9 + \sqrt{3}$$

Therefore, the correct option is C.

Given: $|x^2 - 8x + 15| - 2x + 7 = 0$

Find: The sum of all roots.

First factor the quadratic inside the modulus:

$$x^2 - 8x + 15 = (x - 3)(x - 5)$$

So,

• $x^2 - 8x + 15 \geq 0$ when $x \leq 3$ or $x \geq 5$
• $x^2 - 8x + 15 < 0$ when $3 < x < 5$

Also, from

$$|x^2 - 8x + 15| = 2x - 7$$

the right-hand side must be non-negative. Hence,

$$2x - 7 \geq 0 \Rightarrow x \geq \frac{7}{2}$$

So only the intervals $\frac{7}{2} \leq x < 5$ and $x \geq 5$ need to be checked.

For $x \geq 5$, the expression inside modulus is non-negative, so

$$|x^2 - 8x + 15| = x^2 - 8x + 15$$

Hence,

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

Solving,

$$x = \frac{10 \pm \sqrt{100 - 88}}{2} = \frac{10 \pm \sqrt{12}}{2} = 5 \pm \sqrt{3}$$

Now $5 + \sqrt{3} > 5$, but $5 - \sqrt{3} < 5$. Therefore only $5 + \sqrt{3}$ is valid here.

For $\frac{7}{2} \leq x < 5$, the expression inside modulus is negative, so

$$|x^2 - 8x + 15| = -(x^2 - 8x + 15)$$

Then

$$-(x^2 - 8x + 15) - 2x + 7 = 0$$ $$-x^2 + 6x - 8 = 0$$ $$x^2 - 6x + 8 = 0$$ $$(x - 2)(x - 4) = 0$$

So the roots are $x = 2$ and $x = 4$. Since $\frac{7}{2} \leq x < 5$, only $x = 4$ is valid.

Thus the valid roots are $4$ and $5 + \sqrt{3}$.

The sum of all roots is

$$4 + 5 + \sqrt{3} = 9 + \sqrt{3}$$

Therefore, the correct option is C, that is, $9 + \sqrt{3}$.