Quadratic comparison
Quadratic equation questions in bank exams give two equations and ask how x compares with y. Factorise each by finding two numbers that multiply to the constant and add to the middle coefficient, flip their signs to get the roots, then compare every root of one against every root of the other. If the comparison differs across pairs, no relation can be established.
The quadratic comparison set is five nearly-free marks in a bank prelims paper, and it is the kind where the correct answer is most often the one candidates never select. Papers construct sets where the root ranges overlap so that the honest answer is that no relation holds — and a candidate expecting a clean inequality will pick one anyway.
Try it against the clock. Quadratic Equations runs at 30s easy, 40s medium and 50s hard. The clock is stamped and judged on our server, so the limit you see is the deadline that is actually enforced — and it starts when you tap Start, not while the question is loading.
For x² − 7x + 12, find two numbers multiplying to 12 and adding to −7: that is −3 and −4, so the roots are 3 and 4. Flipping the sign of the factors to get the roots is the step most often fumbled under time pressure.
A positive constant means both factors share the middle term's sign; a negative constant means they have opposite signs and the larger takes the middle term's sign. Knowing which case you are in halves the search immediately.
With roots 3 and 4 for x, and 1 and 2 for y, all four comparisons give x greater than y, so x > y holds. Checking only one pair is how a wrong relation gets selected with full confidence.
If x's roots are 2 and 5 while y's are 3 and 4, then x can be below or above y and no relation can be established. This is a correct and frequently intended answer, not a failure to solve.
| The equation looks like | The factors are |
|---|---|
| x² + bx + c, both positive | Both negative roots |
| x² − bx + c, c positive | Both positive roots |
| x² + bx − c | Opposite signs, larger is negative |
| x² − bx − c | Opposite signs, larger is positive |
| All four comparisons agree | A strict relation holds |
| Comparisons disagree | No relation can be established |
Generated by the same engine that mints the ComputePrep daily. Each answer is computed from the numbers printed in the question, and the walkthrough below each one is the engine's own working — not a solution written afterwards. Reload this page's live drill and you get different numbers.
I. x² + 1x − 2 = 0
II. y² − 5y + 4 = 0
Solve both equations and state the relation between x and y.
I. x² + 11x + 30 = 0
II. y² + 14y + 48 = 0
Solve both equations and state the relation between x and y.
I. x² − 17x + 72 = 0
II. y² − 1y − 12 = 0
Solve both equations and state the relation between x and y.
The wrong options are not random numbers. Each one is the result of a specific careless error for this topic, so picking one tells you which habit is costing you marks.
Factors of −3 and −4 give roots of positive 3 and 4. Reporting the factors as the roots reverses every comparison in the question, and the reversed answer is always an available option.
Both roots of each equation must be checked against both of the other. A relation that holds for one pairing and fails for another means no relation holds at all.
Candidates assume a clean inequality is always intended and pick the closest one. Overlapping root ranges are deliberately constructed, and in those questions “cannot be determined” is the only correct answer.
Find both roots of each equation, then compare every root of the first against every root of the second. If all comparisons point the same way, that relation holds. If they disagree — for instance x's roots straddle y's — then no relation can be established, which is a genuine and common answer.
Use the sign rules first to know what you are looking for, then find two numbers whose product is the constant and whose sum is the middle coefficient. Exam quadratics are constructed with small integer roots, so the search is short once the sign case is settled.
Almost always because the factors were reported as the roots without flipping their signs. Factors of −3 and −4 correspond to roots of 3 and 4. Write the roots on a separate line from the factors and the error largely disappears.
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