Unit 9 · Checkpoint quiz

Unit 9 checkpoint: Quadratics

12 questions · about 25-30 minutes · no time limit

Answer every question, then press Grade my answers at the bottom. You'll see what you got right, a worked solution for each question, and links to the lessons worth revisiting. Pencil and paper are encouraged.

This is practice, not a test of you. A low score just shows where to look next.

Solve \(2(x - 3)^2 = 50\).

Divide both sides by 2: \((x - 3)^2 = 25\).

Square root property (don't forget \(\pm\)): \(x - 3 = \pm 5\).

\(x = 3 + 5 = 8\) or \(x = 3 - 5 = -2\). Answer: \(x = 8\) or \(x = -2\).

Solve \((x + 1)^2 = -9\). Give complex solutions in the form \(a + bi\).

Square root property: \(x + 1 = \pm\sqrt{-9}\).

\(\sqrt{-9} = i\sqrt{9} = 3i\), so \(x + 1 = \pm 3i\).

Subtract 1. Answer: \(x = -1 + 3i\) or \(x = -1 - 3i\).

What number must be added to \(x^2 - 10x\) to make a perfect square trinomial?

Take half of the \(x\)-coefficient: \(\dfrac{-10}{2} = -5\).

Square it: \((-5)^2 = 25\).

Check: \(x^2 - 10x + 25 = (x - 5)^2\). Answer: \(25\).

Solve by completing the square: \(2x^2 + 8x - 6 = 0\).

Divide every term by 2 so the \(x^2\) coefficient is 1: \(x^2 + 4x - 3 = 0\).

Move the constant: \(x^2 + 4x = 3\). Half of 4 is 2, and \(2^2 = 4\); add 4 to both sides: \(x^2 + 4x + 4 = 7\).

Write as a square: \((x + 2)^2 = 7\), so \(x + 2 = \pm\sqrt{7}\).

Answer: \(x = -2 \pm \sqrt{7}\).

Solve with the quadratic formula: \(3x^2 = 2x + 2\).

Standard form first: \(3x^2 - 2x - 2 = 0\), so \(a = 3\), \(b = -2\), \(c = -2\).

Discriminant: \(b^2 - 4ac = (-2)^2 - 4(3)(-2) = 4 + 24 = 28\), and \(\sqrt{28} = 2\sqrt{7}\).

\(x = \dfrac{-(-2) \pm 2\sqrt{7}}{2 \cdot 3} = \dfrac{2 \pm 2\sqrt{7}}{6}\).

Divide every term by 2. Answer: \(x = \dfrac{1 \pm \sqrt{7}}{3}\).

Use the discriminant to describe the solutions of \(2x^2 - 3x + 5 = 0\).

  1. Two non-real complex solutions
  2. Two different real solutions
  3. One repeated real solution
  4. One real, one complex solution

\(a = 2\), \(b = -3\), \(c = 5\).

\(b^2 - 4ac = (-3)^2 - 4(2)(5) = 9 - 40 = -31\).

The discriminant is negative, so the formula has a square root of a negative number: two non-real complex solutions.

Find the vertex of the parabola \(y = 2x^2 - 8x + 3\).

\(x = -\dfrac{b}{2a} = -\dfrac{-8}{2 \cdot 2} = \dfrac{8}{4} = 2\).

\(y = 2(2)^2 - 8(2) + 3 = 8 - 16 + 3 = -5\).

Answer: the vertex is \((2, -5)\).

A ball is thrown upward. Its height after \(t\) seconds is \(h = -16t^2 + 64t + 5\) feet. What is the ball's maximum height, in feet?

The parabola opens down (\(a = -16\)), so the vertex is the highest point.

\(t = -\dfrac{b}{2a} = -\dfrac{64}{2(-16)} = 2\) seconds.

\(h = -16(2)^2 + 64(2) + 5 = -64 + 128 + 5 = 69\). Answer: \(69\) feet.

Solve \(x^2 - x - 12 \lt 0\). Write your answer in interval notation.

Find the zeros: \(x^2 - x - 12 = (x - 4)(x + 3) = 0\), so \(x = 4\) or \(x = -3\).

Test \(x = 0\) (between them): \(0 - 0 - 12 = -12 \lt 0\). True, so the middle interval works.

The parabola opens up, so it is below zero only between the zeros. The sign is \(\lt\), so the ends are not included.

Answer: \((-3, 4)\).

Solve \(x^2 + 2x \ge 15\). Write your answer in interval notation.

Get 0 on one side: \(x^2 + 2x - 15 \ge 0\). Factor: \((x + 5)(x - 3) \ge 0\). Zeros: \(x = -5\) and \(x = 3\).

Test \(x = 0\): \(0 + 0 - 15 = -15\), which is not \(\ge 0\). So the middle fails and the two outside pieces work.

The sign is \(\ge\), so the zeros are included (square brackets).

Answer: \((-\infty, -5] \cup [3, \infty)\).

Solve \(1 + \dfrac{3}{x} = \dfrac{10}{x^2}\).

The LCD is \(x^2\) (and \(x \ne 0\)). Multiply every term by \(x^2\): \(x^2 + 3x = 10\).

Standard form and factor: \(x^2 + 3x - 10 = (x + 5)(x - 2) = 0\), so \(x = -5\) or \(x = 2\).

Neither one makes a denominator 0, so both are solutions. Answer: \(x = 2\) or \(x = -5\).

Factor: \(6x^2 + x - 15\).

\(ac = 6 \cdot (-15) = -90\). Two numbers that multiply to \(-90\) and add to \(1\): \(10\) and \(-9\).

Split the middle term: \(6x^2 + 10x - 9x - 15\).

Group: \(2x(3x + 5) - 3(3x + 5) = (3x + 5)(2x - 3)\).

Answer: \((3x + 5)(2x - 3)\).

Want to look something up first? Quadratics cheat sheet · Glossary

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