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 11 - 2017 - Paper 2

Evaluating the integral gives us:
$\left[ 2x^{\frac{3}{2}} + Ax \right]_0^1 = 2(1) + A(1) - (0) = 2 + A$
This means we can set our original equation to:
$2 + A = 2A^2$

Now, rearranging this equation gives us:
$2A^2 - A - 2 = 0$
This can be solved using the quadratic formula:
$A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Here, (a = 2), (b = -1), and (c = -2).

Calculating the discriminant:
$b^2 - 4ac = (-1)^2 - 4(2)(-2) = 1 + 16 = 17$
Since the discriminant is greater than zero, this indicates that there are two distinct real roots.

Substituting into the quadratic formula gives:
$A = \frac{1 \pm \sqrt{17}}{4}$
Thus, the two possible values for A are:
$A_1 = \frac{1 + \sqrt{17}}{4}, \quad A_2 = \frac{1 - \sqrt{17}}{4}$
This confirms that there are exactly two values for A.