求下列一元二次方程的精确根。

1. Factorise the quadratic expression to its linear factors.

$x^2 + 3x + 2 = 0$
$(x + 1)(x + 2) = 0$

2. Equate each factor to zero to solve for the exact roots.

$x + 1 = 0 \implies x = -1$
$x + 2 = 0 \implies x = -2$

1. Multiply the equation by $-1$ to make the leading coefficient positive for easier factorisation.

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

2. Factorise the quadratic expression.

$(x + 3)(x - 1) = 0$

3. Determine the roots by setting each factor to zero.

$x + 3 = 0 \implies x = -3$
$x - 1 = 0 \implies x = 1$

1. Multiply the entire equation by $-2$ to remove the fraction and make the $x^2$ coefficient positive.

$-\frac{x^2}{2} + 2x - 1 = 0$
$x^2 - 4x + 2 = 0$

2. Apply the quadratic formula to find the roots.

$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(2)}}{2(1)}$
$x = \frac{4 \pm \sqrt{16 - 8}}{2}$

3. Simplify the surd to obtain the exact roots in their simplest form.

$x = \frac{4 \pm \sqrt{8}}{2}$
$x = \frac{4 \pm 2\sqrt{2}}{2}$
$x = 2 \pm \sqrt{2}$

1. Eliminate the fractions by multiplying the equation by the lowest common multiple of the denominators.

$12 \left(\frac{2}{3}x^2 + x - \frac{1}{4}\right) = 0$
$8x^2 + 12x - 3 = 0$

2. Apply the quadratic formula to find the roots.

$x = \frac{-(12) \pm \sqrt{(12)^2 - 4(8)(-3)}}{2(8)}$
$x = \frac{-12 \pm \sqrt{144 + 96}}{16}$
$x = \frac{-12 \pm \sqrt{240}}{16}$

3. Simplify the surd to obtain the exact final roots.

$\sqrt{240} = \sqrt{16 \times 15} = 4\sqrt{15}$
$x = \frac{-12 \pm 4\sqrt{15}}{16}$
$x = \frac{-3 \pm \sqrt{15}}{4}$

1. Rearrange the given equation into the standard quadratic form $ax^2 + bx + c = 0$.

$-\pi x^2 + x + \pi = 1$

$\pi x^2 - x + 1 - \pi = 0$

2. Apply the quadratic formula using $a = \pi$, $b = -1$, and $c = 1 - \pi$.

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

$x = \frac{1 \pm \sqrt{1 - 4\pi + 4\pi^2}}{2\pi}$

3. Recognise the discriminant as a perfect square to simplify the square root.

Observe that $4\pi^2 - 4\pi + 1 = (2\pi - 1)^2$.

$x = \frac{1 \pm \sqrt{(2\pi - 1)^2}}{2\pi}$

$x = \frac{1 \pm (2\pi - 1)}{2\pi}$

4. Evaluate the two cases to find the exact roots.

Case 1:
$x = \frac{1 + (2\pi - 1)}{2\pi} = \frac{2\pi}{2\pi} = 1$

Case 2:
$x = \frac{1 - (2\pi - 1)}{2\pi} = \frac{2 - 2\pi}{2\pi} = \frac{2(1 - \pi)}{2\pi} = \frac{1 - \pi}{\pi}$

1. Substitute the coefficients into the quadratic formula

$a = \frac{1}{2}$, $b = \sqrt{6}$, $c = 3$

$x = \frac{-\sqrt{6} \pm \sqrt{(\sqrt{6})^2 - 4\left(\frac{1}{2}\right)(3)}}{2\left(\frac{1}{2}\right)}$

2. Simplify the discriminant to find the exact root

$x = \frac{-\sqrt{6} \pm \sqrt{6 - 6}}{1}$
$x = -\sqrt{6}$