Given that A is constant and $$\int^1_0 (3\sqrt{x} + A) \, dx = 2A^2$$ show that there are exactly two possible values for A. - Edexcel - A-Level Maths Pure - Question 12 - 2017 - Paper 2

We have established that:

$2 + A = 2A^2$

Rearranging this equation gives:

$2A^2 - A - 2 = 0$

To find the possible values for A, we can use the quadratic formula:

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

Where in our case, (a = 2), (b = -1), and (c = -2). Substituting these values in:

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

This simplifies to:

$A = \frac{1 \pm \sqrt{1 + 16}}{4} = \frac{1 \pm \sqrt{17}}{4}$

Thus, we find two possible values for A:

$A_1 = \frac{1 + \sqrt{17}}{4}, \quad A_2 = \frac{1 - \sqrt{17}}{4}$

The results show that there are exactly two values for A, confirming the statement. Therefore, the values of A are:

$A = \frac{1 + \sqrt{17}}{4} \quad \text{and} \quad A = \frac{1 - \sqrt{17}}{4}$