OMPT drill — Algebra

OMPT quadratic equations practice: lesson and 10 problems

Q: How do I answer a discriminant question with a parameter on the OMPT?

Compute D = b^2 - 4ac with the parameter left in, then set the condition the question names: D = 0 for exactly one solution, D > 0 for two, D < 0 for none. Solve that for the parameter and keep every value: p^2 = 36 gives p = 6 and p = -6. If the parameter sits in front of x^2, also check the value that makes it zero.

Q: How should I type roots and fractions as quadratic answers on the OMPT?

Exactly, not as decimals. Fractions go in as 1/2, roots as sqrt(5), so 2 + sqrt(5) is typed as 2+sqrt(5). Simplify first: sqrt(20) becomes 2sqrt(5). Enter both solutions when there are two. When the discriminant is negative, answer "no solution". Decimals are only right when the question says to round.

Q: Are quadratic equations on the OMPT-C?

Only lightly. The OMPT-C has no separate quadratic chapter; its four chapters are numbers, algebra, linear formulas and equations, and geometry. Quadratics show up inside the algebra chapter as factoring and expanding, and occasionally as a simple equation with no middle term. If you are preparing for OMPT-C, spend your time on factoring and on linear equations rather than on the discriminant.

Q: What is the difference between a quadratic equation and a quadratic inequality on the OMPT?

An equation asks where ax^2 + bx + c equals zero and the answer is up to two numbers. An inequality asks where it is above or below zero and the answer is a region, such as -2 < x < 3. You find the same roots either way, then for the inequality you sketch the parabola and read off which side of the roots is wanted. Both appear on the OMPT.

Factoring, completing the square, and the abc-formula, plus knowing which one saves you a minute. The discriminant question format turns up on nearly every variant. Work through all ten below before moving on; algebra slips compound into every other strand.

Tested inOMPT-A OMPT-B OMPT-C OMPT-D OMPT-E OMPT-F OMPT-G

OMPT quadratic equations: the lesson

What the OMPT tests in quadratic equations

A quadratic equation has the shape $a x^{2} + b x + c = 0$ , and the squared term changes the game: there can be two solutions, one, or none at all. Everything in this topic is about finding them fast and knowing in advance how many to expect. You have three tools, factoring, completing the square, and the abc-formula, and the OMPT quietly tests whether you pick the right one, because the right tool takes thirty seconds and the wrong one takes three minutes.

The chapter "Quadratic equations" is on every variant except that the OMPT-C only touches it through its algebra chapter. The heavy sub-skills are the discriminant with a parameter ("for which $p$ has the equation exactly one solution"), factoring when the numbers are friendly, and the formula when they are not. Lighter but regular: equations with no middle term, a common factor of $x$ , and story problems about areas or thrown balls. My rule of thumb from years of marking: try to factor for about ten seconds. If two numbers that multiply to $c$ and add to $b$ do not jump out, stop hunting and use the formula. The formula always works; factoring is a shortcut, not a duty.

Notation and answer format the OMPT expects

Give every solution. Two roots means two entries, " $x = 2$ or $x = 3$ ". A question that wants one of them says so: "the smaller solution", "the sum of the solutions".
Roots stay exact: type $2 + sqrt (5)$ and $2 - sqrt (5)$ , never $4.24$ . Simplify the root: $20$ becomes $25$ .
Fractions stay fractions: $1/2$ , not $0.5$ , unless the question says "round to".
If the discriminant is negative, the answer is "no solution" (no real solution). The OMPT does not use complex numbers.
A discriminant question with a parameter usually has two answers for the parameter too, such as $p = 6$ or $p = - 6$ .

Worked example 1: factoring when it is quick

Solve $x^{2} - 5 x + 6 = 0$ .
I need two numbers multiplying to $+ 6$ and adding to $- 5$ . Both negative, then: $- 2$ and $- 3$ .
So $x^{2} - 5 x + 6 = (x - 2) (x - 3) = 0$ .
A product is zero only when a factor is zero, so $x = 2$ or $x = 3$ .
Substituting back gives $4 - 10 + 6 = 0$ and $9 - 15 + 6 = 0$ . Both solutions go in the answer box; the trap is stopping after the first one.

Worked example 2: the formula when it is not

Solve $2 x^{2} + 3 x - 2 = 0$ . With $a = 2$ the mental factoring is slower, so go straight to the formula.
Compute the discriminant first, on its own line: $D = 3^{2} - 4 \cdot 2 \cdot (- 2) = 9 + 16 = 25$ . Positive, and a perfect square: two rational solutions coming.
Then $x = \frac{- 3 \pm 5}{4}$ , giving $x = \frac{2}{4} = \frac{1}{2}$ or $x = \frac{- 8}{4} = - 2$ .
Check the ugly one: $2 \cdot \frac{1}{4} + \frac{3}{2} - 2 = \frac{1}{2} + \frac{3}{2} - 2 = 0$ . Good.
The trap sits inside $D$ : with $c = - 2$ , the term $- 4 a c$ is $+ 16$ , not $- 16$ . Writing $D$ on its own line is how you catch that.

The full lesson adds the discriminant with a parameter, irrational roots and how to simplify them, the common-factor and no-middle-term shapes, equations that only become quadratic after a rewrite, a word problem, and a "how many solutions" question of the hardest kind the OMPT sets. It ends with the eight slips I mark most often.

How each OMPT variant tests quadratic equations

The chapter weights are the official ones from the omptest.org learning-assessment pages. The question counts are my estimates of how much of that chapter this one topic takes up. The 21-question papers (OMPT-D and OMPT-F) ask long multi-part questions, so a topic there usually shows up as a step inside a bigger problem.

OMPT-A
52 questions
Testedroughly 5 to 7 of 52 questions
Official chapter: Quadratic equations and functions, 22% of the paper.
Numeric answers: solve by factoring or the formula, a discriminant question with a parameter, and one area or motion word problem.
OMPT-B
64 questions
Testedroughly 5 to 8 of 64 questions
Official chapter: Quadratic equations and functions, 10% of the paper.
Same as OMPT-A plus hidden quadratics: an equation with the unknown in a denominator, a substitution in $x^{2}$ , and quadratics that finish a trig or calculus item.
OMPT-C
30 questions
Testedroughly 1 to 3 of 30 questions
Official chapter: Algebra, 25% of the paper.
No separate chapter; quadratics appear inside the algebra chapter as factoring and expanding, with an occasional simple equation to solve.
OMPT-D
21 questions
Tested1 to 2 of the 21 long questions, usually as a step inside a bigger problem
Official chapter: Functions, equations and inequalities, 30% of the paper.
The discriminant with a parameter, or a quadratic that decides where a curve meets a line, as one part of a long question.
OMPT-E
40 questions
Testedroughly 3 to 5 of 40 questions
Official chapter: Quadratic equations and functions, 10% of the paper.
Short numeric items: factor, use the formula, count solutions with the discriminant. Word problems about area or revenue appear.
OMPT-F
21 questions
Tested1 to 2 of the 21 long questions, usually as a step inside a bigger problem
Official chapter: Equations and inequalities, 30% of the paper.
Multi-part functions questions where a quadratic gives the intersection points or the zeros, sometimes with a parameter to be found from the discriminant.
OMPT-G
26 questions
Testedroughly 2 to 4 of 26 questions
Official chapter: Functions, graphs and equations, 45% of the paper.
Quadratics from physics: a projectile height, a braking distance, a kinematics formula solved for time. Reject the root that has no physical meaning.

OMPT quadratic equations practice problems

10 original problems in the formats the OMPT uses: typed numeric answers, multiple choice, and spot-the-error. The last few are harder than the exam average on purpose. Solutions and the common mistake sit behind each problem.

Problem 1Numeric answer

Solve $w^{2} - 7 w + 12 = 0$ . Give the smaller of the two solutions.

Show the worked solution

Answer: 3

I look for two numbers that multiply to 12 and add to $- 7$ : those are $- 3$ and $- 4$ . So the equation factors as $(w - 3) (w - 4) = 0$ , giving $w = 3$ or $w = 4$ . The smaller solution is 3.

Problem 2Multiple choice

For which value of $c$ does $2 y^{2} + 8 y + c = 0$ have exactly one solution?

A $c = 8$
B $c = - 8$
C $c = 4$
D $c = 16$

Show the worked solution

Answer: A — $c = 8$

Exactly one solution means the discriminant is zero. Here $D = 8^{2} - 4 \cdot 2 \cdot c = 64 - 8 c$ . Setting $64 - 8 c = 0$ gives $c = 8$ . With that value the equation becomes $2 (y + 2)^{2} = 0$ , a single repeated root at $y = - 2$ .

Problem 3Numeric answer

Solve $3 n^{2} = 12 n$ . Give the nonzero solution.

Show the worked solution

Answer: 4

I bring everything to one side first: $3 n^{2} - 12 n = 0$ , then factor out $3 n$ to get $3 n (n - 4) = 0$ . That gives $n = 0$ or $n = 4$ , and the nonzero solution is 4. I never divide both sides by $n$ — that throws away the $n = 0$ root and on a real test that lost root is often exactly what a follow-up part asks about.

Problem 4Multiple choice

The equation $(t + 2)^{2} = 25$ has two solutions. What is their sum?

A $- 4$
B $4$
C $10$
D $- 10$

Show the worked solution

Answer: A — $- 4$

Taking square roots of both sides gives $t + 2 = 5$ or $t + 2 = - 5$ , so $t = 3$ or $t = - 7$ . Their sum is $3 + (- 7) = - 4$ . You can also see this without solving: both roots sit symmetrically around $t = - 2$ , so they must add to $- 4$ .

Problem 5Spot the error

A student solves $x^{2} - 5 x = 6$ like this. Step 1: factor the left side as $x (x - 5) = 6$ . Step 2: so $x = 6$ or $x - 5 = 6$ . Step 3: solutions $x = 6$ and $x = 11$ . Which step contains the error?

Show the worked solution

Answer: Step 2

The factoring in Step 1 is legal, but Step 2 misuses the zero-product rule: from $A \cdot B = 6$ you cannot conclude $A = 6$ or $B = 6$ . That rule only works when the product equals zero. The right move is $x^{2} - 5 x - 6 = 0$ , which factors as $(x - 6) (x + 1) = 0$ , giving $x = 6$ and $x = - 1$ . The student's $x = 6$ happens to be correct by coincidence, but $x = 11$ fails: $121 - 55 = 66 \neq = 6$ .

Problem 6Numeric answer

A rectangular vegetable bed has an area of $40 m^{2}$ . Its length is 3 metres more than its width. Find the width in metres.

Show the worked solution

Answer: 5

With width $b$ the length is $b + 3$ , so $b (b + 3) = 40$ , which rearranges to $b^{2} + 3 b - 40 = 0$ . I factor: $(b + 8) (b - 5) = 0$ , so $b = - 8$ or $b = 5$ . A width cannot be negative, so the bed is 5 metres wide (and $5 \times 8 = 40$ confirms it).

Problem 7Numeric answer

Solve $2 x^{2} - 7 x - 15 = 0$ . Give the larger of the two solutions.

Show the worked solution

Answer: 5

With $a = 2$ I go straight to the discriminant: $D = (- 7)^{2} - 4 \cdot 2 \cdot (- 15) = 49 + 120 = 169 = 1 3^{2}$ . Then $x = \frac{7 \pm 13}{4}$ , so $x = \frac{20}{4} = 5$ or $x = \frac{- 6}{4} = - \frac{3}{2}$ . The larger solution is 5. Check: $2 \cdot 25 - 35 - 15 = 50 - 50 = 0$ . The factored form is $(2 x + 3) (x - 5)$ , which says the same thing.

Problem 8Multiple choice

What are the solutions of $x^{2} + 6 x + 4 = 0$ ?

A $x = - 3 \pm 5$
B $x = - 3 \pm 25$
C $x = 3 \pm 5$
D $x = - 6 \pm 5$

Show the worked solution

Answer: A — $x = - 3 \pm 5$

No integer pair multiplies to 4 and adds to 6, so I use the formula. $D = 36 - 16 = 20$ , and $20 = 25$ . So $x = \frac{- 6 \pm 2 5}{2}$ , and dividing every term by 2 gives $x = - 3 \pm 5$ . Check with $x = - 3 + 5$ : $x^{2} = 14 - 65$ , and $x^{2} + 6 x + 4 = 14 - 65 - 18 + 65 + 4 = 0$ .

Problem 9Numeric answer

The equation $x^{2} - k x + 2 k - 3 = 0$ has exactly one solution for two different values of $k$ . Give the sum of those two values.

Show the worked solution

Answer: 8

Exactly one solution means $D = 0$ . Here $a = 1$ , $b = - k$ , $c = 2 k - 3$ , so $D = k^{2} - 4 (2 k - 3) = k^{2} - 8 k + 12$ . Setting it to zero gives a quadratic in $k$ : $k^{2} - 8 k + 12 = (k - 2) (k - 6) = 0$ , so $k = 2$ or $k = 6$ . Their sum is 8. Check $k = 2$ : $x^{2} - 2 x + 1 = (x - 1)^{2}$ , one root. Check $k = 6$ : $x^{2} - 6 x + 9 = (x - 3)^{2}$ , one root.

Problem 10Numeric answer

A photo measuring 10 cm by 15 cm is surrounded by a frame of uniform width $w$ cm. The photo and frame together have an area of $266 cm^{2}$ . Find $w$ .

Show the worked solution

Answer: 2

The frame adds $w$ on every side, so the outer rectangle is $(10 + 2 w)$ by $(15 + 2 w)$ . Its area: $(10 + 2 w) (15 + 2 w) = 150 + 50 w + 4 w^{2} = 266$ . Bring to zero and halve: $4 w^{2} + 50 w - 116 = 0$ , so $2 w^{2} + 25 w - 58 = 0$ . Discriminant: $D = 625 + 464 = 1089 = 3 3^{2}$ . Then $w = \frac{- 25 \pm 33}{4}$ , giving $w = 2$ or $w = - \frac{29}{2}$ . A width cannot be negative, so $w = 2$ . Check: $14 \times 19 = 266$ .

OMPT quadratic equations: questions students ask

Are quadratic equations on the OMPT-A?

Yes. "Quadratic equations" is one of the eight OMPT-A chapters, so roughly 5 to 7 of the 52 questions come from it. Expect an equation to factor, one that needs the abc-formula, a discriminant question with a parameter asking for exactly one or no solution, and a short word problem about an area or a thrown object. Quadratic functions and their graphs are a separate chapter.

Do I need the abc-formula for the OMPT, or is factoring enough?

You need both. Factoring is faster when two integers multiply to $c$ and add to $b$ , but the OMPT sets equations such as $x^{2} - 4 x - 1 = 0$ whose roots are $2 \pm 5$ , and those only come out of the formula. My rule: hunt for factors for ten seconds, then switch. Write the discriminant on its own line every time.

How do I answer a discriminant question with a parameter on the OMPT?

Compute $D = b^{2} - 4 a c$ with the parameter left in, then set the condition the question names: $D = 0$ for exactly one solution, $D > 0$ for two, $D < 0$ for none. Solve that for the parameter and keep every value: $p^{2} = 36$ gives $p = 6$ and $p = - 6$ . If the parameter sits in front of $x^{2}$ , also check the value that makes it zero.

How should I type roots and fractions as quadratic answers on the OMPT?

Exactly, not as decimals. Fractions go in as 1/2, roots as sqrt(5), so $2 + 5$ is typed as 2+sqrt(5). Simplify first: $20$ becomes $25$ . Enter both solutions when there are two. When the discriminant is negative, answer "no solution". Decimals are only right when the question says to round.

Are quadratic equations on the OMPT-C?

Only lightly. The OMPT-C has no separate quadratic chapter; its four chapters are numbers, algebra, linear formulas and equations, and geometry. Quadratics show up inside the algebra chapter as factoring and expanding, and occasionally as a simple equation with no middle term. If you are preparing for OMPT-C, spend your time on factoring and on linear equations rather than on the discriminant.

What is the difference between a quadratic equation and a quadratic inequality on the OMPT?

An equation asks where $a x^{2} + b x + c$ equals zero and the answer is up to two numbers. An inequality asks where it is above or below zero and the answer is a region, such as $- 2 < x < 3$ . You find the same roots either way, then for the inequality you sketch the parabola and read off which side of the roots is wanted. Both appear on the OMPT.

Test hubs that cover quadratic equations:OMPT-A OMPT-B OMPT-C OMPT-D OMPT-E OMPT-F OMPT-G All 28 topics

OMPT quadratic equations practice: lesson and 10 problems

OMPT quadratic equations: the lesson

What the OMPT tests in quadratic equations

Notation and answer format the OMPT expects

Worked example 1: factoring when it is quick

Worked example 2: the formula when it is not

How each OMPT variant tests quadratic equations

OMPT quadratic equations practice problems

OMPT quadratic equations: questions students ask

Related OMPT topics and test hubs