Title: random choice
Course: Phil's test course

Q: This is a Numerical question.

Solve $$x^2 + 3x - 28 = 0$$

Enter either of the possible answers.
A: -7 / 4
Feedback ("-7"): Your answer is correct.
Feedback ("4"): Your answer is correct.
Feedback (any other answer): Your answer is incorrect.
Feedback: The quadratic equation is $$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ a = 1, b = 3 and c = -28. $$x = \frac{-3 \pm \sqrt{3^2 - 4\times 1\times(-28)}}{2\times a}$$ $$x = \frac{-3 \pm \sqrt{9 + 112}}{2}$$ $$x = \frac{-3 \pm \sqrt{121}}{2}$$ $$x = \frac{-3 \pm 11}{2}$$ $$x = \frac{-14}{2}$$ or $$\frac{8}{2}$$ and the solutions are x = -7 and x = 4. We can prove these are correct by substituting the values into the original equation. $$(-7)^2 + 3\times(-7) - 28 = 49 - 21 - 28 = 0$$ $$4^2 + 3\times4 - 28 = 16 + 12 -28 = 0$$
  Hint: You may need to be reminded of the quadratic formula $$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
  Hint: You will need to substitute for a, b and c in the quadratic formula $$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ a is the coefficient for $$x^2$$, b is coefficient for $$x$$ and c is the constant.

Q: This is an OU multiple response question.

Which four of the following expressions are equivalent to $$4y$$?
- $$\frac{2y+6y}{2}$$
- $$4\frac{y^2}{y}$$
- $$\frac{1}{2}(2y+6y)$$
- $$2(y+y)$$
- $$2y+\frac{6y}{2}$$
- $$\frac{2y^2}{y}$$
- $$\sqrt{16}y^2$$
- $$6(y-2)$$
A: $$\frac{2y+6y}{2}$$ / $$4\frac{y^2}{y}$$ / $$\frac{1}{2}(2y+6y)$$ / $$2(y+y)$$
Feedback: The following four expressions may all be simplified to $$4y$$: $$\frac{2y+6y}{2}$$ = $$\frac{8y}{2}$$ = $$4y$$; $$4\frac{y^2}{y}$$ = $$4y$$ $$\frac{1}{2}(2y+6y)$$ = $$\frac{1}{2}8y$$ = $$4y$$; $$2(y+y)$$ = $$2(2y)$$ = $$4y$$; The other four expressions simplify to give the following: $$2y+\frac{6y}{2}$$ = $$2y+3y$$ = $$5y$$; $$\frac{2y^2}{y}$$ = $$2y$$ $$\sqrt{16}y^2$$ = $$4y^2$$; $$6(y-2)$$ = $$6y - 12$$
  Hint: Start by expanding all the expressions to remove the brackets.
  Hint: For your final attempt any incorrect choices that you have made will be removed.

Q: This is a Multiple choice question.

Estimate the gradient of the curve shown in the plot at x = 2.

[Image: Curve]

Click on the answer closest to your estimate.
- 1 (This value is too low.)
- 2 (This value is too low.)
- 3 (This value is too low.)
- 4 (The gradient is indeed 4.)
- 5 (This value is too high.)
- 6 (This value is too high.)
A: 4
Feedback: [Image: Curve with tangent] Gradient of tangent = $$\frac{\Delta y}{\Delta x} = \frac{14 - 0}{4.0 - 0.5} = \frac{14}{3.5} = 4.0$$ The gradient of a curve at any point is the same as the gradient of a tangent drawn at that point, so the gradient of the curve is 4.0 at x = 2.
  Hint: Start by imagining a tangent drawn against the curve at x = 2. Your tangent should be exactly parallel to the curve at x = 2.
  Hint: Where does your tangent, drawn at x = 2, cross the x and y axes? Estimate these points then calculate the gradient as the difference in y values divided by the difference in x values.