Solve for x: 1.1 x(5x+2)=0; 1.2 x(2x-3)=4 (correct to TWO decimal places); 1.3 x2-x-6 >= 6; 1.4 2^x+2 + 2^x+2 = 42; 1.5 sqrt(4x-11) = 2-x

Question 1 (a)

Step 1: Apply the quadratic formula
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
where $a = 3$, $b = -8$, $c = 4$.

Step 2: Compute the discriminant
$\Delta = b^2 - 4ac = (-8)^2 - 4 \cdot 3 \cdot 4 = 64 - 48 = 16.$

Step 3: Take the square root
$\sqrt{\Delta} = \sqrt{16} = 4.$

Step 4: Substitute the values
$x = \frac{8 \pm 4}{2 \cdot 3} = \frac{8 \pm 4}{6}.$

Step 5: Simplify
$x = \frac{8 + 4}{6} = \frac{12}{6} = 2,$
$x = \frac{8 - 4}{6} = \frac{4}{6} = \frac{2}{3}.$

x = 2,\ x = \dfrac{2{3}}

Question 1 (b)

Step 1: Apply the quadratic formula
$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
where $a = 4$, $b = -8$, $c = -3$.

Step 2: Compute the discriminant
$\Delta = b^2 - 4ac = (-8)^2 - 4 \cdot 4 \cdot (-3) = 64 + 48 = 112.$

Step 3: Simplify the discriminant
$\Delta = 16 \cdot 7, \quad \sqrt{\Delta} = 4\sqrt{7}.$

Step 4: Substitute the values
$y = \frac{8 \pm 4\sqrt{7}}{2 \cdot 4} = \frac{8 \pm 4\sqrt{7}}{8} = \frac{2 \pm \sqrt{7}}{2}.$

y = \dfrac{2 + \sqrt{7}{2},\ y = \dfrac{2 - \sqrt{7}}{2}}

Question 2 (a)

Step 1: From $x - y = 1$, solve for $y$
$y = x - 1.$

Step 2: Substitute into the first equation
$\sqrt{x^2 + (x - 1)^2} = 3.$

Step 3: Square both sides
$x^2 + (x - 1)^2 = 9.$

Step 4: Expand
$x^2 + (x^2 - 2x + 1) = 9$
$2x^2 - 2x + 1 = 9$
$2x^2 - 2x - 8 = 0.$

Step 5: Divide by 2
$x^2 - x - 4 = 0.$

Step 6: Apply the quadratic formula
$x = \frac{1 \pm \sqrt{(-1)^2 - 4(1)(-4)}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 + 16}}{2} = \frac{1 \pm \sqrt{17}}{2}.$

Step 7: Find corresponding $y$ values
$y = x - 1 = \frac{1 \pm \sqrt{17}}{2} - 1 = \frac{1 \pm \sqrt{17} - 2}{2} = \frac{-1 \pm \sqrt{17}}{2}.$

\left( \dfrac{1 + \sqrt{17}{2},\ \dfrac{-1 + \sqrt{17}}{2} \right),\ \left( \dfrac{1 - \sqrt{17}}{2},\ \dfrac{-1 - \sqrt{17}}{2} \right)}

Question 2 (c)

Step 1: Set $y = 0$ to find roots
$x^2 - 3x + 2 = 0.$

Step 2: Factor the quadratic
$(x - 1)(x - 2) = 0.$

Step 3: Solve
$x = 1,\ x = 2.$

The graph is a parabola opening upwards with vertex at $\left( \frac{3}{2}, -\frac{1}{4} \right)$ and two distinct real roots.

x=1,\ x=2