SSC CGL Quadratic Equation Practice Test 2026 — Free Chapter Wise Questions

Quadratic Equation Tests — Pick a Level and Start

Tap a level to see every test in it. Start at Level 1 as soon as you can read the sum and product of the roots straight off the coefficients, then move to Level 2 with the timer running — that is where two relations get chained inside one question, exactly as the real paper does it.

Quadratic equation is the chapter where algebra stops being manipulation and starts being reading. Almost every question in it can be answered from two numbers — the sum of the roots and their product — and almost every mistake in it comes from finding the roots when they were never needed. A candidate who internalises that one habit answers most of the chapter in under forty seconds a question.

It is also unusually good value per hour. The sum-and-product relations reappear in the algebraic identity chapters, the discriminant reappears whenever a question asks whether something is possible, and the factorising practice pays out in simplification and in data-sufficiency style framings. These practice tests put twenty-five quadratic questions in front of you in one sitting, each with a full solution and the shortcut, in English or Hindi.

How Many Quadratic Equation Questions Are Asked in SSC CGL?

Counted directly it is a one-to-two question chapter in Tier I. Counted with the places its relations are reused, it sits under a noticeably larger share of the algebra block — which is why it repays study out of proportion to its own mark count.

The third and fourth rows are the ones candidates underrate. Dividing an equation through by x to reach x + 1/x is a quadratic-equation move, and so is deciding whether a pair of lines meets.

What You Will Practice in the Quadratic Equation Chapter

Sum and product of the roots — the two numbers that answer everything

For ax² + bx + c = 0 with roots α and β, the sum is −b/a and the product is c/a. Every symmetric expression in the roots is reachable from those two alone, which is why the roots themselves are almost never worth finding.

Two of these are worth memorising as sentences rather than formulas. The gap between the roots squared IS the discriminant divided by a², so any question about how far apart the roots are is a discriminant question in disguise. And reciprocal roots mean the product is 1, which happens exactly when c = a.

The discriminant and the nature of the roots

D = b² − 4ac decides everything about what kind of roots exist, before any of them is found.

• D > 0 — two distinct real roots. If D is also a perfect square, they are rational and the expression factorises over whole numbers.
• D = 0 — the two roots merge into one repeated root, and the expression is a perfect square such as (2x − 3)². A zero discriminant is easy to misread as “no roots”; it means the opposite.
• D < 0 — no real roots. The sum and product relations still hold and still give real answers, which is why a question can ask for α² + β² even here.

When a parameter sits inside the coefficients, the equal-roots condition D = 0 becomes an equation in that parameter. If the parameter appears squared you get two values and both must be reported; if the squared terms cancel you get one. Read which case you are in before assuming the answer count.

Signs and position of the roots

The product fixes the signs and the sum separates the remaining cases. A negative product means one root of each sign, and nothing else needs checking — a negative product also guarantees the roots are real. A positive product with a positive sum means both are positive; a positive product with a negative sum means both are negative. In each of those two cases the discriminant must still be tested separately, and it frequently trims the range a careless answer would allow.

For questions about where the roots sit relative to a number t, use the curve. With a positive leading coefficient the expression is negative strictly between its roots, so f(t) < 0 means t lies between them. To place both roots beyond t you need three conditions at once: D ≥ 0, f(t) > 0, and the turning point −b/2a on the correct side of t.

Building the equation whose roots are transformed

Find the new sum and the new product from the old ones, then write x² − (new sum)x + (new product) = 0 and clear any fractions at the end. The original roots are never needed.

The reciprocal row has a shortcut worth knowing: reversing the coefficients of the original equation produces the reciprocal-root equation directly, so 2x² − 7x + 3 = 0 becomes 3x² − 7x + 2 = 0. Note also that the shifted product is not P + h² — the cross term hS is the piece most often dropped.

Common roots

If two quadratics share ONE root, subtract them. The shared root satisfies both and therefore satisfies the difference, and the x² terms cancel to leave a linear equation that yields the root at once. Scale one equation first if the leading coefficients differ. The subtraction gives a candidate only, so substitute it back to find the unknown.

If they share BOTH roots, the two equations describe the same pair of numbers and can differ only by an overall scaling, so their coefficients are in proportion: a₁/a₂ = b₁/b₂ = c₁/c₂. Reading which of the two situations the question describes is half the work.

Factor theorem, remainder theorem and equations in disguise

If (x − p) is a factor of a polynomial then the polynomial is zero at x = p; a stated remainder means it equals that remainder instead. The same substitution answers both. When the divisor is (mx + n) the test value is the fraction −n/m, and when the divisor is a quadratic the remainder is of the form rx + s, which needs two substitutions to pin down.

A large family of questions is quadratic only after one move: a fractional equation cleared of denominators, a radical equation squared, a quartic in x², a repeated block replaced by a single letter, or an equation divided through by x to reach x + 1/x. In every one of these the last step is the one candidates forget — converting back, and discarding any value that breaks the original equation.

Consistency of linear equations in two variables

Compare the three ratios of a₁x + b₁y = c₁ and a₂x + b₂y = c₂:

• a₁/a₂ ≠ b₁/b₂ — the lines cross once, so there is exactly one solution whatever the constants are.
• a₁/a₂ = b₁/b₂ = c₁/c₂ — the same line twice, so infinitely many solutions.
• a₁/a₂ = b₁/b₂ ≠ c₁/c₂ — parallel and distinct, so no solution at all.

The constants are what separate the last two cases, so the third ratio is never a formality. When a parameter appears in several places, solve with two ratios and then TEST the answer in the third — a candidate that passes the first pair can still fail there.

Maximum and minimum of a quadratic expression

Complete the square: x² + bx + c = (x + b/2)² + (c − b²/4). A square is never negative, so with a positive leading coefficient the leftover constant is the minimum, reached at x = −b/2a; with a negative leading coefficient it is the maximum. Factor the leading coefficient out before completing the square, and read the question carefully — the position x and the value are different answers and both are offered.

Solved Examples — SSC CGL Quadratic Equation

How the Quadratic Equation Chapter Tests Work

• 25 questions per test, with a full question palette — answered, skipped, marked for review, not visited.
• Optional 15-minute timer — the same sectional clock as the real paper, with it to build speed and without it while you are still learning the relations.
• Marking: +2 correct, −0.5 wrong, 0 unattempted — the same scheme as the real paper.
• Solutions, shortcut tricks and concept notes open after you submit, question by question.
• English and हिंदी, chosen at the start of every test.
• Free with a TrickySSC account — sign in once with Google or your mobile number, and every test is open to you.

How to Prepare Quadratic Equation for SSC CGL

1. Write the sum and the product first, every single time. Before reading the rest of the question, put S = −b/a and P = c/a on the page. A large share of this chapter is answered from those two lines alone.
2. Stop solving for the roots. If the question is symmetric in α and β, the roots are a detour. Finding them is the single biggest time loss in the chapter.
3. Learn the four identities cold — S² − 2P, S² − 4P, S³ − 3PS and S/P. Everything else in the chapter is built by combining them.
4. Check where the parameter sits before counting answers. A parameter in the constant term usually gives one value; a parameter that ends up squared usually gives two. Reporting one when there are two is a common half-mark loss.
5. Treat the discriminant as a range tool, not just a yes/no test. Questions asking for which values of k the roots are real, distinct, equal or of a given sign are discriminant questions with an inequality at the end.
6. For common roots, subtract; for both roots common, use proportion. Two different situations, two different methods, and the question always tells you which.
7. Undo every substitution. Whether you set y = x², t = x + 1/x or named a repeated block, the answer asked for is in x. Handing back the helper variable is the most frequent error in the disguised-equation family.
8. Do Level 1 untimed first. The aim is the correct relation, not speed. Move to Level 2 with the 15-minute clock on — that is 36 seconds a question, the real Tier I pace.
9. Verify against a real paper. Once Level 2 is consistent, attempt a previous year paper and then a full mock test under locked sectional timing.

Related SSC CGL Maths Chapters

These chapters share the algebra that this one is built on. Doing them next keeps the same toolkit in play.

All SSC CGL Chapter Wise Tests

Every chapter below has its own free Level 1 and Level 2 tests with full solutions.

SSC CGL Quadratic Equation — Frequently Asked Questions

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