2.3 Quadratic Equations & Quantity Comparisons (Q1 vs Q2)

2.3 Quadratic Equations & Quantity Comparisons (Q1 vs Q2)

Quadratic Equations and Quantity Comparisons form a high-scoring core block in the Quantitative Aptitude paper. In SBI PO Prelims, candidates encounter 5 quadratic equation questions requiring them to solve two equations in variables $x$ and $y$ and determine their mathematical relationship. In SBI PO Mains, this concept evolves into Quantity Comparison (Quantity I vs Quantity II), where candidates evaluate two independent mathematical expressions or arithmetic word problems to establish an inequality.

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The Five Standard Answer Options

For both Quadratic Equations and Quantity Comparisons, options follow a standard 5-choice format:

1. $x > y$ (or Quantity I > Quantity II)
2. $x < y$ (or Quantity I < Quantity II)
3. $x \ge y$ (or Quantity I $\ge$ Quantity II)
4. $x \le y$ (or Quantity I $\le$ Quantity II)
5. $x = y$ or Relationship cannot be established (CND)

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The Master Sign Shortcut Table

For any quadratic equation in standard form $ax^2 + bx + c = 0$, the signs of the linear coefficient $b$ and constant term $c$ determine the signs of the final roots ($x_1, x_2$) prior to numerical factorization.

The Golden Instant-Answer Rule (Both Constants Negative)

If both equations have negative constant terms ($c_1 < 0$ and $c_2 < 0$):

• Equation I ($x$): One positive root ($+x_1$) and one negative root ($-x_2$).
• Equation II ($y$): One positive root ($+y_1$) and one negative root ($-y_2$).

When comparing:

• $+x_1$ is greater than $-y_2$.
• $-x_2$ is smaller than $+y_1$.

Because $x$ is simultaneously greater than $y$ in one pairing and smaller than $y$ in another, the relationship CANNOT be established. When $c_1 < 0$ and $c_2 < 0$, mark option 5 instantly without doing any calculations!

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Step-by-Step Fast Factorization & Comparison Algorithm

To solve $ax^2 + bx + c = 0$ in under 15 seconds:

1. Product Step: Compute $P = a \times c$.
2. Split Step: Identify two factors $f_1$ and $f_2$ such that $f_1 \times f_2 = P$ and $f_1 + f_2 = b$.
3. Invert Sign Step: Reverse the signs of factors to get ($-f_1, -f_2$).
4. Divide Coefficient Step: Divide by leading coefficient $a$ to get final roots: $x_1 = -f_1 / a$, $x_2 = -f_2 / a$.

Cross-Multiplication Trick to Avoid Fraction Division

If Equation I has leading coefficient $a_1$ and Equation II has leading coefficient $a_2$:

• Instead of dividing $x$-factors by $a_1$ and $y$-factors by $a_2$, simply multiply $x$-factors by $a_2$ and multiply $y$-factors by $a_1$.
• Comparing integers is faster and completely eliminates division errors.

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Worked Example 1: Full Quadratic Comparison

Problem: Compare roots for:

• Equation I: $2x^2 - 11x + 15 = 0$
• Equation II: $3y^2 - 13y + 14 = 0$

Solving Equation I ($x$):

• $a_1 = 2, b_1 = -11, c_1 = 15$. Product $P = 2 \times 15 = 30$.
• Find factors of 30 adding to $-11$: $-6$ and $-5$.
• Invert signs: $+6$ and $+5$.
• Raw roots for $x$: $x_1 = 6/2 = 3$, $x_2 = 5/2 = 2.5$.

Solving Equation II ($y$):

• $a_2 = 3, b_2 = -13, c_2 = 14$. Product $P = 3 \times 14 = 42$.
• Find factors of 42 adding to $-13$: $-7$ and $-6$.
• Invert signs: $+7$ and $+6$.
• Raw roots for $y$: $y_1 = 7/3 \approx 2.33$, $y_2 = 6/3 = 2$.

Constructing Comparative Matrix:

• Compare $x = 3$ against $y = 2.33 \implies x > y$
• Compare $x = 3$ against $y = 2 \implies x > y$
• Compare $x = 2.5$ against $y = 2.33 \implies x > y$
• Compare $x = 2.5$ against $y = 2 \implies x > y$

All 4 pairings confirm $x > y$. Final Answer: $x > y$.

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Worked Example 2: Radical / Irrational Coefficient Equations

In SBI PO Mains, equations feature radical linear terms (e.g., $x^2 - 7\sqrt{3}x + 36 = 0$).

Step-by-step Reduction Method:

1. Divide the constant term $c$ by the integer inside the square root sign. Here, divide $36$ by $3 = 12$.
2. Find factors of $12$ that add up to the coefficient outside the radical (7). Factors of 12 adding to 7 are $4$ and $3$.
3. Attach the square root $\sqrt{3}$ to both factors: $4\sqrt{3}$ and $3\sqrt{3}$.
4. Invert signs based on sign rules (since $-b, +c$, roots are positive). Roots of $x$: $x_1 = 4\sqrt{3} \approx 6.93$, $x_2 = 3\sqrt{3} \approx 5.20$.

Now compare with Equation II ($y^2 - 5\sqrt{2}y + 12 = 0$):

• Divide $12$ by $2 = 6$.
• Factors of 6 summing to 5 are $3$ and $2$.
• Roots of $y$: $y_1 = 3\sqrt{2} \approx 4.24$, $y_2 = 2\sqrt{2} \approx 2.83$.

Comparing: $x$ roots ($\{5.20, 6.93\}$) are strictly greater than $y$ roots ($\{2.83, 4.24\}$). Final Answer: $x > y$.

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Quantity Comparison Framework (Quantity I vs Quantity II)

Quantity Comparison questions in Mains replace abstract equations with complete mini-word problems.

Worked Example 3: Arithmetic Quantity Comparison

Problem:

• Quantity I: The speed of Train A (in km/h) which passes a telegraph pole in 10 seconds and has a length of 200 meters.
• Quantity II: The speed of Train B (in km/h) which passes a platform of length 100 meters in 20 seconds, given the length of Train B is 300 meters.

Evaluating Quantity I:

$$\text{Speed of Train A} = \frac{\text{Distance}}{\text{Time}} = \frac{200 \text{ m}}{10 \text{ s}} = 20 \text{ m/s}$$

Convert m/s to km/h by multiplying by $18/5$:

$$\text{Quantity I} = 20 \times \frac{18}{5} = 72 \text{ km/h}$$

Evaluating Quantity II:

$$\text{Total Distance} = \text{Length of Train B} + \text{Length of Platform} = 300 + 100 = 400 \text{ meters}$$ $$\text{Speed of Train B} = \frac{400 \text{ m}}{20 \text{ s}} = 20 \text{ m/s}$$ $$\text{Quantity II} = 20 \times \frac{18}{5} = 72 \text{ km/h}$$

Comparison: $\text{Quantity I} = \text{Quantity II} = 72 \text{ km/h}$. Final Answer: Quantity I = Quantity II.

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Top 3 Common Traps in Quadratic Equations

Trap 1: Square Root vs. Quadratic Power

• Equation A: $x^2 = 144 \implies x = +12$ and $x = -12$ (Two roots).
• Equation B: $y = \sqrt{144} \implies y = +12$ only (Square root symbol denotes principal positive root).
• Result: When $x = +12$, $x = y$; when $x = -12$, $x < y$. Final relation is $x \le y$.

Trap 2: Variable Cancellation Loss

• Incorrect: $x^2 = 4x \implies \text{Divide by } x \implies x = 4$.
• Correct: $x^2 - 4x = 0 \implies x(x - 4) = 0 \implies x = 0$ or $x = 4$.
• Canceling a variable eliminates the root $0$, causing incorrect comparison conclusions.

Trap 3: Inequality Sign Reversal

When multiplying or dividing an inequality by a negative value, the direction of the inequality flips ($x > y \implies -x < -y$).