### Related Formula
For equations with repeating polynomial expressions, applying a variable substitution like
t = P(x)$t = P(x)$ reduces high-degree polynomials down to standard quadratics.
### Core Logic
Expand the linear binomial product in the given equation equation:
(x-4)(x-5) = x^2 - 9x + 20$$(x-4)(x-5) = x^2 - 9x + 20$$
Rewrite the full expression in terms of the common variable pattern
x^2 - 9x$x^2 - 9x$:
(x^2 - 9x + 11)^2 - (x^2 - 9x + 20) = 3$$(x^2 - 9x + 11)^2 - (x^2 - 9x + 20) = 3$$
Let
t = x^2 - 9x$t = x^2 - 9x$. Substituting this into the equation gives:
(t + 11)^2 - (t + 20) = 3$$(t + 11)^2 - (t + 20) = 3$$
t^2 + 22t + 121 - t - 20 - 3 = 0$$t^2 + 22t + 121 - t - 20 - 3 = 0$$
t^2 + 21t + 98 = 0$$t^2 + 21t + 98 = 0$$
### Step 1: Solve the Polynomial for Variable t
Factorize the quadratic expression:
(t + 14)(t + 7) = 0 implies t = -14 text or t = -7$$(t + 14)(t + 7) = 0 \implies t = -14 \text{ or } t = -7$$
### Step 2: Back-substitute and Isolate Rational Roots
Case 1:
x^2 - 9x = -7 implies x^2 - 9x + 7 = 0$x^2 - 9x = -7 \implies x^2 - 9x + 7 = 0$
Check the discriminant value:
D = (-9)^2 - 4(1)(7) = 81 - 28 = 53$D = (-9)^2 - 4(1)(7) = 81 - 28 = 53$ (not a perfect square, so the roots are irrational).
Case 2:
x^2 - 9x = -14 implies x^2 - 9x + 14 = 0$x^2 - 9x = -14 \implies x^2 - 9x + 14 = 0$
Factorize the quadratic expression:
(x - 7)(x - 2) = 0 implies x = 7 text or x = 2$$(x - 7)(x - 2) = 0 \implies x = 7 \text{ or } x = 2$$
Both values are rational numbers.
### Step 3: Calculate the
Product of Rational Roots
Multiply the true rational roots together:
textProduct = 7 cdot 2 = 14$$\text{Product} = 7 \cdot 2 = 14$$
### Pattern Recognition
Always check the discriminant
D = b^2 - 4ac$D = b^2 - 4ac$ to filter out irrational radical components whenever the problem specifically asks for the product of *rational* roots only.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Mathematics: Quadratic Equations
