Answer all questions.

1. Substitute the given values of $a$ and $c$ into the equation

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

2. Square both sides and solve for $b$

$16 = \frac{b^2 - 16}{8}$
$128 = b^2 - 16$
$b^2 = 144$
$b = \pm 12$

1. Simplify the first bracket by applying the negative fractional index

$\left(\frac{125p^6}{q^{-3}}\right)^{-\frac{1}{3}} = \left(125p^6q^3\right)^{-\frac{1}{3}} = 125^{-\frac{1}{3}}p^{6(-\frac{1}{3})}q^{3(-\frac{1}{3})} = \frac{1}{5}p^{-2}q^{-1}$

2. Simplify the second bracket using the zero index and squaring

$\left(\frac{3p^0}{\sqrt{q}}\right)^2 = \left(\frac{3(1)}{q^{\frac{1}{2}}}\right)^2 = \frac{3^2}{q^{\frac{1}{2} \times 2}} = \frac{9}{q}$

3. Multiply the simplified terms and express with positive indices

$\left(\frac{1}{5}p^{-2}q^{-1}\right) \times \left(\frac{9}{q}\right) = \frac{9}{5}p^{-2}q^{-2} = \frac{9}{5p^2q^2}$

1. Align the coefficients of $y$ to prepare for elimination.

Let $2x - 3y = -15$ be equation (1).
Let $0.6x + 1.5y = 6$ be equation (2).
Multiply equation (2) by 2:
$1.2x + 3y = 12$ --- (3)

2. Eliminate $y$ and solve for $x$.

Add equation (1) and equation (3):
$(2x - 3y) + (1.2x + 3y) = -15 + 12$
$3.2x = -3$
$x = -\frac{3}{3.2} = -\frac{15}{16} = -0.9375$

3. Substitute the value of $x$ back to find $y$.

Substitute $x = -\frac{15}{16}$ into equation (1):
$2\left(-\frac{15}{16}\right) - 3y = -15$
$-\frac{15}{8} - 3y = -15$
$3y = 15 - \frac{15}{8}$
$3y = \frac{105}{8}$
$y = \frac{35}{8} = 4.375$