Question 2

Let's solve each quadratic equation by finding its roots.

Step 1 — Solve $x^2 - 3x - 10 = 0$

We will factor the quadratic expression. We need two numbers that multiply to -10 and add to -3. These numbers are -5 and 2.

$x^2 - 5x + 2x - 10 = 0$

$x(x - 5) + 2(x - 5) = 0$

$(x - 5)(x + 2) = 0$

This means either $x - 5 = 0$ or $x + 2 = 0$.

$\boxed{x = 5 \text{ or } x = -2}$

Step 2 — Solve $2x^2 + x - 6 = 0$

We will factor this quadratic equation. We look for two numbers whose product is $2 \times (-6) = \textbf{-12}$ and whose sum is 1. These numbers are 4 and -3.

$2x^2 + 4x - 3x - 6 = 0$

$2x(x + 2) - 3(x + 2) = 0$

$(2x - 3)(x + 2) = 0$

This means either $2x - 3 = 0$ or $x + 2 = 0$.

$\boxed{x = \frac{3}{2} \text{ or } x = -2}$

Step 3 — Solve $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$

We will use the quadratic formula to find the roots. The coefficients are $a = \sqrt{2}$, $b = 7$, and $c = 5\sqrt{2}$.

First, let's calculate the discriminant $D = b^2 - 4ac$.

$D = (7)^2 - 4(\sqrt{2})(5\sqrt{2})$

$D = 49 - 4(5 \times 2)$

$D = 49 - 40$

$D = 9$

Now, we apply the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$.

$x = \frac{-7 \pm \sqrt{9}}{2\sqrt{2}}$

$x = \frac{-7 \pm 3}{2\sqrt{2}}$

We find the two possible roots.

$x_1 = \frac{-7 + 3}{2\sqrt{2}}$

$x_1 = \frac{-4}{2\sqrt{2}}$

$x_1 = \frac{-2}{\sqrt{2}}$

$x_1 = -\sqrt{2}$

$x_2 = \frac{-7 - 3}{2\sqrt{2}}$

$x_2 = \frac{-10}{2\sqrt{2}}$

$x_2 = \frac{-5}{\sqrt{2}}$

$x_2 = \frac{-5\sqrt{2}}{2}$

$\boxed{x = -\sqrt{2} \text{ or } x = \frac{-5\sqrt{2}}{2}}$

Step 4 — Solve $2x^2 - x + 1 + \frac{1}{8} = 0$

First, let's simplify the constant terms.

$2x^2 - x + \frac{8}{8} + \frac{1}{8} = 0$

$2x^2 - x + \frac{9}{8} = 0$

Let's multiply the entire equation by 8 to remove the fraction.

$8(2x^2) - 8(x) + 8\left(\frac{9}{8}\right) = 0$

$16x^2 - 8x + 9 = 0$

Now, we use the quadratic formula. The coefficients are $a = 16$, $b = -8$, and $c = 9$.

Let's calculate the discriminant $D = b^2 - 4ac$.

$D = (-8)^2 - 4(16)(9)$

$D = 64 - 576$

$D = -512$

Since the discriminant $D$ is negative, there are no real roots for this equation.

$\boxed{\text{No real roots}}$

Step 5 — Solve $100x^2 - 20x + 1 = 0$

We can recognize this as a perfect square trinomial. It is in the form $(Ax - B)^2$.

We can write $100x^2$ as $(10x)^2$ and $1$ as $(1)^2$. The middle term is $-20x$, which is $-2(10x)(1)$.

$(10x)^2 - 2(10x)(1) + (1)^2 = 0$

$(10x - 1)^2 = 0$

This means $10x - 1$ must be equal to 0.

$10x - 1 = 0$

$10x = 1$

$\boxed{x = \frac{1}{10}}$

Answer

(i) $x = 5$ or $x = -2$ (ii) $x = \frac{3}{2}$ or $x = -2$ (iii) $x = -\sqrt{2}$ or $x = \frac{-5\sqrt{2}}{2}$ (iv) No real roots (v) $x = \frac{1}{10}$