Quadratic Equations — IBPS Clerk / SBI Clerk

Banking-exam quadratic questions follow a distinctive format rarely seen elsewhere: two separate quadratic equations are given, one in $x$ and one in $y$, and the question asks for the relationship between $x$ and $y$ — not simply the roots of either equation alone. Solving both equations correctly is only the first half of the question; comparing every possible pairing of roots is the second, equally important half.

1. Solving a quadratic equation by factoring

Almost every banking-exam quadratic factors cleanly into two integer roots — find two numbers that multiply to the constant term and add to the coefficient of the linear term (with correct signs), then split the equation into two factors. $x^2-5x+6=0$ needs two numbers multiplying to $6$ and adding to $-5$: those are $-2$ and $-3$, giving $(x-2)(x-3)=0$, so $x=2$ or $x=3$.

2. Sum and product of roots (a quick verification)

For any quadratic $a{x}^2+bx+c=0$, the sum of the roots equals $-\frac{b}{a}$ and the product equals $\frac{c}{a}$ — a fast way to verify factored roots without re-expanding the factored form.

$\text{Sum of roots}=-\frac{b}{a},\text{Product of roots}=\frac{c}{a}$

For $x^2-5x+6=0$: sum should be $5$ and product should be $6$ — and indeed $2+3=5$ and $2\times 3=6$, confirming the factoring.

3. The Quantity I vs Quantity II format

Once both equations are solved, EVERY root of the first equation must be compared against EVERY root of the second equation — since a variable's "value" is genuinely any one of its roots, not a single fixed number, until every combination is checked. A two-root equation compared against a two-root equation therefore has up to 4 combinations to check, not just 1.

4. Reading the outcome across all combinations

If every combination consistently shows $x\ge y$ (with at least one combination giving equality and none giving $x<y$), the answer is "$x\ge y$"; if every combination consistently shows $x>y$ with no equality anywhere, the answer is the strict "$x>y$"; and if different combinations give conflicting relationships (some $x>y$, some $x<y$), the relationship CANNOT be determined.

5. Why checking only one combination is a common trap

Checking only the "obvious" pairing — say, the larger root of each equation — and stopping there is the single most common error, since the relationship must hold across EVERY combination to be a valid conclusion, not just the one combination that happened to be checked first. A relationship that looks consistent for one pairing can reverse entirely for another, which is exactly why the "cannot be determined" outcome exists as a legitimate answer.

Worked Examples

Example 1. Solve: $x^2-5x+6=0$ and $y^2-7y+12=0$. Compare $x$ and $y$.

Equation 1 factors as $(x-2)(x-3)=0$, so $x=2$ or $x=3$. Equation 2 factors as $(y-3)(y-4)=0$, so $y=3$ or $y=4$.

Checking all 4 combinations: $(2,3)\to x<y$; $(2,4)\to x<y$; $(3,3)\to x=y$; $(3,4)\to x<y$. Every combination gives $x<y$ or $x=y$, never $x>y$.

Answer: $x\le y$.

Example 2. Solve: $x^2-4x-21=0$ and $y^2-2y-24=0$. Compare $x$ and $y$.

Equation 1 factors as $(x-7)(x+3)=0$, so $x=7$ or $x=-3$. Equation 2 factors as $(y-6)(y+4)=0$, so $y=6$ or $y=-4$.

Checking all 4 combinations: $(7,6)\to x>y$; $(7,-4)\to x>y$; $(-3,6)\to x<y$; $(-3,-4)\to x>y$. This mixes $x>y$ and $x<y$ across different combinations.

Answer: Relationship cannot be determined.

Example 3. Solve: $x^2-6x+9=0$ and $y^2-10y+21=0$. Compare $x$ and $y$.

Equation 1 factors as $(x-3{)}^2=0$, giving a single repeated root $x=3$. Equation 2 factors as $(y-3)(y-7)=0$, so $y=3$ or $y=7$.

Checking both combinations: $(3,3)\to x=y$; $(3,7)\to x<y$. Every combination gives $x=y$ or $x<y$, never $x>y$.

Answer: $x\le y$.

Example 4. Verify the roots of $x^2-9x+20=0$ using the sum-and-product check.

Factoring: two numbers multiplying to 20 and adding to 9 are 4 and 5, giving $(x-4)(x-5)=0$, so $x=4$ or $x=5$. Verification: sum should be $9$ (indeed $4+5=9$) and product should be $20$ (indeed $4\times 5=20$).

Answer: $x=4$ or $x=5$, verified.

Summary

Banking-exam quadratic questions give two separate equations — one in $x$, one in $y$ — and ask for the relationship between the variables, not just the roots of either equation alone.

Most banking-exam quadratics factor cleanly: find two numbers multiplying to the constant term and adding to the linear coefficient (with correct signs). The sum-and-product check ($-b/a$ and $c/a$) verifies factored roots quickly without re-expanding.

Every root of the first equation must be compared against every root of the second — a two-root-vs-two-root comparison has up to 4 combinations. The relationship is only valid if it holds CONSISTENTLY across every combination; checking just one combination and stopping is the most common error.

If every combination agrees ($x>y$ throughout, or $x<y$ throughout, or mixed with equality but never the opposite strict inequality), a definite relationship ($x>y$, $x<y$, $x=y$, $x\ge y$, or $x\le y$) can be stated. If combinations genuinely conflict, "relationship cannot be determined" is the correct — and legitimate — answer.