The sum of all real values of \( x \) for which \[ \frac{3x^2 - 9x + 17}{x^2 + 3x + 10} = \frac{5x^2 - 7x + 19}{3x^2 + 5x + 12} \] is equal to \_\_\_\_\_.

1. Why clear denominators?

Whenever you see

with polynomials in a JEE question, the standard first move is to cross-multiply:

Doing so removes the fractions and turns the problem into one long polynomial equation.

2. Expanding: keeping track of terms

After cross-multiplying, you must multiply out carefully (FOIL, or distributive law):

1. Multiply the highest-degree terms to get the $x^4$ term.
2. Multiply cross-terms for $x^3$, $x^2$, $x$, and the constant. A small slip here creates a wrong coefficient and ruins the rest of the work, so be systematic.

3. Collect like terms → obtain a single polynomial

Once expanded, bring everything to the left side so the equation reads

Grouping like powers of $x$ (all $x^4$ together, all $x^3$ together, …) gives a clean quartic.

4. Factoring a quartic: why and how

A quartic $ax^4+bx^3+cx^2+dx+e=0$ rarely needs the full Ferrari method in JEE.
Examiners usually design it so that it factors into quadratics with integer coefficients.
Common hints:

• The leading coefficient and the constant often share small factors (like 2 and 7 here).
• The constant term is small, so the product of the last constants in each quadratic is easy to guess.

By trying $(2x^2+px+q)(x^2+rx+s)$ and matching coefficients, you can crack it quickly.

5. Real roots only

Once the quartic splits, check each quadratic’s discriminant $\Delta = b^2-4ac$.

• $\Delta>0$ → two real roots.
• $\Delta=0$ → one real root (double).
• $\Delta<0$ → no real roots (ignore for the sum).

6. Safety check: denominators must not vanish

Original denominators $x^2+3x+10$ and $3x^2+5x+12$ must never be zero.
Happily, both have negative discriminants, so they’re positive for all real $x$ and no root is lost here.

7. Sum of roots — shortcut

If you only need the sum of a quadratic’s roots, use Vieta: for $ax^2+bx+c=0$, the sum is $-\frac{b}{a}$.
That lets you avoid even punching the square-root button!

What does the question want?

Two bulky fractions (made from polynomials) are told to be perfectly equal.
Your job: find every real number $x$ that makes this true and then add them together.

How can we think of it like a 10-year-old?

Imagine two see-saws. Each see-saw has a funny plank whose weight depends on $x$.
If the planks balance exactly, you have found a good $x$. Collect all such good $x$’s and just add them like pocket-money coins.

Baby steps to reach the answer

1. Clear the fractions – like wiping away the denominators so you only have a big “normal” equation.
2. Expand & tidy up – open all brackets so you get a single polynomial.
3. Factor – break that big polynomial into smaller pieces (just like breaking a Lego tower into blocks).
4. Roots – each block gives possible $x$’s.
5. Check – be sure the original fractions don’t explode (denominator never zero).
6. Add – finally, put the good roots together and add.

That’s it!

Step 1: Cross-multiply

Step 2: Expand both sides

Left side:

Right side:

Step 3: Bring to one side and simplify

Divide by 2:

Step 4: Factor the quartic

Assume $(2x^2+px+q)(x^2+rx+s)=0$ with small integers. Setting up coefficient matching gives a consistent set when $q = 7,\; s = 1,\; p = -12,\; r = 1.$ Thus

Step 5: Inspect each quadratic

1. $x^2 + x + 1 = 0$
   $\Delta = 1 - 4 = -3 < 0$ → no real roots.
2. $2x^2 - 12x + 7 = 0$
   $x = \dfrac{12 \pm \sqrt{144 - 56}}{4} = \dfrac{12 \pm \sqrt{88}}{4}$
   Two real roots: $$x_1 = \frac{12 + \sqrt{88}}{4}, \qquad x_2 = \frac{12 - \sqrt{88}}{4}$$

Step 6: Sum of all real roots

Using Vieta on $2x^2 -12x + 7$ (or simply add $x_1$ and $x_2$):

Step 7: Denominator check

$\;x^2 + 3x + 10$ and $3x^2 + 5x + 12$ have negative discriminants, so they are never zero for real $x$.
Therefore both $x_1$ and $x_2$ are valid.

Final Answer: 6