Solve for x: 1.1.1 x(x−3)=0 1.1.2 2x2+1=4x 1.1.3 x2−2x−3>0 1.1.4 22x-2^x+2−32=0 1.1.5 sqrt(-2x+4)−x=2. Solve for x and y simultaneously: 2x+y=3, y2+xy=2

$$x = 0 \quad \text{or} \quad x = 3$$

The solutions for $x$ are:

$$\boxed{x=0 \text{ or } x=3}$$

1.1.2 Solve for $x$: $2x^2+1=4x$ (correct to TWO decimal places)

Here, $a=2$, $b=-4$, $c=1$.

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

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

$$x_1 = \frac{2 + \sqrt{2}}{2} \approx \frac{2 + 1.41421}{2} \approx \frac{3.41421}{2} \approx 1.7071 \approx 1.71$$ $$x_2 = \frac{2 - \sqrt{2}}{2} \approx \frac{2 - 1.41421}{2} \approx \frac{0.58579}{2} \approx 0.2929 \approx 0.29$$

The solutions for $x$ are:

$$\boxed{x \approx 1.71 \text{ or } x \approx 0.29}$$

1.1.3 Solve for $x$: $x^2 - 2x - 3 > 0$

Factor the quadratic expression:

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

This gives critical points $x=3$ and $x=-1$.

Since the parabola $y=x^2-2x-3$ opens upwards (coefficient of $x^2$ is positive), the expression is greater than zero outside its roots.
We test values in the intervals $(-\infty, -1)$, $(-1, 3)$, and $(3, \infty)$.

• For $x < -1$ (e.g., $x=-2$): $(-2)^2 - 2(-2) - 3 = 4 + 4 - 3 = 5 > 0$ (True)
• For $-1 < x < 3$ (e.g., $x=0$): $(0)^2 - 2(0) - 3 = -3 < 0$ (False)
• For $x > 3$ (e.g., $x=4$): $(4)^2 - 2(4) - 3 = 16 - 8 - 3 = 5 > 0$ (True) The solution for $x$ is:

$$\boxed{x < -1 \text{ or } x > 3}$$

1.1.4 Solve for $x$: $2^{2x} - 2^{x+2} - 32 = 0$

$$2^{2x} = (2^x)^2$$ $$2^{x+2} = 2^x \cdot 2^2 = 4 \cdot 2^x$$

Substitute these into the equation:

$$(2^x)^2 - 4 \cdot 2^x - 32 = 0$$

Substitute $k$ into the equation to form a quadratic equation.

$$k^2 - 4k - 32 = 0$$

$$(k-8)(k+4) = 0$$

This gives two possible values for $k$:

$$k = 8 \quad \text{or} \quad k = -4$$

Case 1: $2^x = 8$

$$2^x = 2^3$$ $$x = 3$$

Case 2: $2^x = -4$
An exponential function with a positive base, like $2^x$, can never be negative. Therefore, there is no real solution for $x$ in this case.
The solution for $x$ is:

$$\boxed{x=3}$$

1.1.5 Solve for $x$: $\sqrt{-2x+4} - x = 2$

$$(\sqrt{-2x+4})^2 = (x+2)^2$$ $$-2x+4 = x^2 + 4x + 4$$

$$0 = x^2 + 4x + 2x + 4 - 4$$ $$0 = x^2 + 6x$$

$$x(x+6) = 0$$

This gives two possible solutions:

$$x = 0 \quad \text{or} \quad x = -6$$

For $x=0$:

$$\sqrt{-2(0)+4} - 0 = 2$$ $$\sqrt{4} - 0 = 2$$ $$2 = 2$$

This solution is valid.
For $x=-6$:

$$\sqrt{-2(-6)+4} - (-6) = 2$$ $$\sqrt{12+4} + 6 = 2$$ $$\sqrt{16} + 6 = 2$$ $$4 + 6 = 2$$ $$10 = 2$$

This statement is false, so $x=-6$ is an extraneous solution.
The solution for $x$ is:

$$\boxed{x=0}$$

2 Solve for $x$ and $y$ simultaneously:
$2x + y = 3 \quad \text{(Equation 1)}$
$y^2 + xy = 2 \quad \text{(Equation 2)}$

$$(9 - 12x + 4x^2) + (3x - 2x^2) = 2$$ $$9 - 12x + 4x^2 + 3x - 2x^2 = 2$$ $$2x^2 - 9x + 9 = 2$$ $$2x^2 - 9x + 7 = 0$$

$$(2x-7)(x-1) = 0$$

This gives two possible values for $x$:

$$2x-7=0 \Rightarrow x = \frac{7}{2}$$ $$x-1=0 \Rightarrow x = 1$$

Case 1: If $x = \frac{7}{2}$

$$y = 3 - 2\left(\frac{7}{2}\right)$$ $$y = 3 - 7$$ $$y = -4$$

Case 2: If $x = 1$

$$y = 3 - 2(1)$$ $$y = 3 - 2$$ $$y = 1$$

The pairs of solutions $(x,y)$ are:

$$\boxed{\left(\frac{7}{2}, -4\right) \text{ and } (1, 1)}$$