The first three terms of a sequence are: Term 1: 6x - 3 Term 2: x² - 2x Term 3: 4x² + 3x (a) Fill in the following table to write each of the first differences in its simplest form in terms of x - Junior Cycle Mathematics - Question 14 - 2019

For First Difference 2:

$$First Difference 2 = 4x^2 + 3x - (x^2 - 2x) = 4x^2 + 3x - x^2 + 2x = 3x^2 + 5x$$

For the sequence to be linear, the first differences must be constant. This means:

$$T_2 - T_1 = T_3 - T_2 \Rightarrow x^2 - 8x + 3 = 3x^2 + 5x$$

Now, simplifying this gives:

$$0 = 3x^2 + 5x - (x^2 - 8x + 3) = 3x^2 + 5x - x^2 + 8x - 3 = 2x^2 + 13x - 3$$

Using the quadratic formula:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Where:

• a = 2,
• b = 13,
• c = -3.

Substituting these values gives:

$$x = \frac{-13 \pm \sqrt{(13)^2 - 4(2)(-3)}}{2(2)} = \frac{-13 \pm \sqrt{169 + 24}}{4} = \frac{-13 \pm \sqrt{193}}{4}$$

Calculating further, the rounded solutions are:

• $x \approx -0.223$ (to three decimal places)
• $x \approx -6.723$ (to three decimal places)