1.1 Solve for $x$: 1.1.1 $(3x-6)(x+2)=0$ 1.1.2 $2x^2-6x+1=0$ (correct to TWO decimal places) 1.1.3 $x^2-90>x$ 1.1.4 $x-7=rac{x}{ ext{√x}}=-12$ 1.2 Solve for $x$ and $y$ simultaneously: $2x-y=2$ $xy=4$ 1.3 Show that $2.5^n - 5^n + 5^2$ is even for all positive integer values of $n$ - NSC Mathematics - Question 1 - 2022 - Paper 1

Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 2$, $b = -6$, $c = 1$:

1. Determine the discriminant:

   $b^2 - 4ac = (-6)^2 - 4(2)(1) = 36 - 8 = 28$

2. Find the roots:

   $x = \frac{6 \pm \sqrt{28}}{4} \\ x = \frac{6 \pm 2\sqrt{7}}{4} \\ x = \frac{3 \pm \sqrt{7}}{2}$

Calculating this gives:

• $x \approx 2.82$
• $x \approx 0.18$

Rearranging the inequality gives:

$x^2 - x - 90 > 0$

1. Finding the critical points:

   $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}=\frac{1 \pm \sqrt{1 + 4 \cdot 90}}{2} = \frac{1 \pm 19}{2}$

Thus, the critical points are $x = 10$ and $x = -9$.

2. We test intervals $(-\infty, -9)$, $(-9, 10)$, and $(10, \infty)$, leading to:
   • For $x < -9$, the inequality holds.
   • For $-9 < x < 10$, it does not.
   • For $x > 10$, the inequality holds.

Final solution: $x < -9$ or $x > 10$.

Isolating $ext{√x}$:

1. Start with:

$x - 7 = \frac{x}{\sqrt{x}} \\ x - 7 = \sqrt{x}$

2. Squaring both sides:

   $(x - 7)^2 = x$

   This simplifies to:

   $x^2 - 14x + 49 = x \\ x^2 - 15x + 49 = 0$

3. Solving this using the quadratic formula leads to:

   $x = \frac{15 \pm \sqrt{15^2 - 4\cdot1\cdot49}}{2 ext{.}1}$

   • Roots are $x = 9$ and $x = 16$.

Using substitution:

1. From $2x - y = 2$, express $y$:

$y = 2x - 2$

2. Substitute into $xy = 4$:

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

3. Solving gives:

   $x = \frac{1 \pm \sqrt{1 + 8}}{2} \\ x = 2, x = -1$

4. For $x = 2$, $y = 2(2) - 2 = 2$.
   For $x = -1$, $y = 2(-1) - 2 = -4$.

Hence, solutions are $(x=2,y=2)$ and $(x=-1,y=-4)$.

1. Simplifying the expression:

   $2.5^n - 5^n + 25 = 2.5^n - (5^n - 25)$

2. Knowing that $5^n$ is odd:

   • $2 imes ext{(odd)}$ is even. Thus, the expression simplifies:

   $(2.5^n) - (5^n - 25)$

   Since both $5^n$ and $25$ are odd, their difference is even, confirming our result.

Thus, the expression is even for all positive integers $n$.

1. Begin by simplifying both sides:

   The left side is:

   $\frac{3^{1/2}}{32} = \frac{\sqrt{3}}{32}$

2. For the right side, simplifying $ext{√96}$ gives:

   $\sqrt{96} = \sqrt{16\cdot 6} = 4\sqrt{6}$

3. Equating both sides:

   $\frac{\sqrt{3}}{32} = \frac{4\sqrt{6}}{1}$

   From here, we would need to analyze to find possible values for $x$ and $y$.

   Thus, as they represent relationships or ratios, we would typically propose values based on these simplifications, confirming the correctness based on the equations first proposed.