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. Cross-multiplication – getting rid of denominators

For two rational expressions

$\frac{P_1(x)}{Q_1(x)} = \frac{P_2(x)}{Q_2(x)}$

the standard move is:

$P_1(x)\,Q_2(x) = P_2(x)\,Q_1(x)$

because multiplying both sides by the product $Q_1\,Q_2$ clears the fractions.

2. Expand carefully

After cross-multiplying we obtain a single polynomial equality. Expanding (multiplying out) is mechanical but you must keep terms in correct order:

1. Multiply each term of the first quadratic with each term of the second.
2. Collect like powers of $x$.

3. Set everything to zero → get a polynomial equation

Once both sides are polynomials, bring all terms to one side. We end up with a quartic (degree-4) polynomial:

$2x^4 - 10x^3 - 3x^2 - 5x + 7 = 0$

4. Factorisation insight

Instead of using the brutal quartic formula, we try to split it into two quadratics:

$\bigl(x^2 + ax + b\bigr)\bigl(2x^2 + cx + d\bigr)$

Matching coefficients (comparing powers of $x$) is a common JEE technique. Clues:

• The constant term is $7$, so $b\,d = 7$ → limited factor pairs (1,7) or (-1,-7).
• Integers often work; test $(b,d)=(1,7)$ first.

Systematically solving the coefficient-matching system quickly gives

$x^2 + x + 1 \quad \text{and} \quad 2x^2 - 12x + 7$

5. Discriminant check for real roots

• For $x^2 + x + 1 = 0$, $\Delta = 1 - 4 = -3 < 0$ → no real roots.
• For $2x^2 - 12x + 7 = 0$, $\Delta = (-12)^2 - 4\times2\times7 = 144 - 56 = 88 > 0$ → two real roots.

6. Sum of real roots

From Vieta’s formula, the sum of roots of $2x^2 - 12x + 7 = 0$ is $\frac{-(-12)}{2} = 6$.

Thus the answer is 6.

What is the problem?

We have two big fractions made of polynomials and we are asked to find all values of $x$ that make both sides exactly equal, and then add those values together.

How to think about it like a 10-year-old?

1. Imagine two balanced see-saws – one fraction on the left, one on the right. We want to know which positions (values of $x$) keep the see-saw perfectly balanced.
2. Clear the fractions by cross-multiplying (just like multiplying both sides of an equality by the denominators so that the sticks under the see-saw vanish). Now we only have a single big polynomial equation – no more fractions!
3. Break the big monster into smaller monsters by trying to factor the polynomial. Sometimes it splits into two smaller quadratics (like breaking a chocolate bar at the grooves).
4. Check which smaller monsters give real roots (because sometimes the roots are imaginary, which means they are not on our real number line playground).
5. Add the real roots you find. That sum is the answer.

Step-by-step solution

1. Start equation

$\frac{3x^2 - 9x + 17}{x^2 + 3x + 10} = \frac{5x^2 - 7x + 19}{3x^2 + 5x + 12}$

2. Cross-multiply

$(3x^2 - 9x + 17)(3x^2 + 5x + 12) = (5x^2 - 7x + 19)(x^2 + 3x + 10)$

3. Expand both sides

Left side: $9x^4 - 12x^3 + 42x^2 - 23x + 204$

Right side: $5x^4 + 8x^3 + 48x^2 - 13x + 190$

4. Bring everything to one side

$9x^4 - 12x^3 + 42x^2 - 23x + 204 - \bigl(5x^4 + 8x^3 + 48x^2 - 13x + 190\bigr) = 0$

Simplify: $4x^4 - 20x^3 - 6x^2 - 10x + 14 = 0$

Divide by $2$: $2x^4 - 10x^3 - 3x^2 - 5x + 7 = 0$

5. Factor the quartic

Guessing a split into two quadratics: $\bigl(x^2 + x + 1\bigr)\bigl(2x^2 - 12x + 7\bigr) = 0$

6. Solve each quadratic

(a) $x^2 + x + 1 = 0$ → $\Delta = -3$ (no real roots).

(b) $2x^2 - 12x + 7 = 0$

$x = \frac{12 \pm \sqrt{88}}{4} = 3 \pm \frac{\sqrt{22}}{2}$

7. Sum of real roots

Using Vieta on $2x^2 - 12x + 7$, the sum is

$\frac{-(-12)}{2} = 6$

Hence, the required sum is 6.