Three bags are shown in the table below - Junior Cycle Mathematics - Question 12 - 2016

To find the values of y where two of the bags have the same mass, we need to set the expressions equal to each other.

1. Set the first two bags equal:

   $y + 5 = 19$

   Solving for y:

   $y = 14$

2. Set the second and third bags equal:

   $19 = 2y^2 + 1$

   Rearranging gives:

   $2y^2 = 18$ $y^2 = 9$ $y = 3$ (positive root)

3. Now, set the first and third bags equal:

   $y + 5 = 2y^2 + 1$

   Rearranging results in:

   $2y^2 - y - 4 = 0$

   Applying the quadratic formula: $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where a = 2, b = -1, c = -4:

   Solving:

   $y = \frac{1 \pm \sqrt{(-1)^2 - 4(2)(-4)}}{2(2)}$

   $y = \frac{1 \pm \sqrt{1 + 32}}{4}$

   $y = \frac{1 \pm \sqrt{33}}{4}$

   The two solutions are:

   $y_1 = \frac{1 + \sqrt{33}}{4}$ $y_2 = \frac{1 - \sqrt{33}}{4}$ (only $y_1$ is valid since y must be positive)

Thus, the three positive values of y are:

• $14$
• $3$
• $\frac{1 + \sqrt{33}}{4} \approx 1.69$ (correct to two decimal places)