OMPT drill — Algebra

Quadratic equations

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 six 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

Lesson

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.

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 $b$ (when $a = 1$ ) do not jump out, stop hunting and use the formula. Students lose far more time forcing a factorisation that does not exist over integers than they ever lose writing out the formula. The formula always works; factoring is a shortcut, not a duty.

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).