Question: Find the range of values of $m$ for which $y = (m - 3)x^2 + mx + m$ lies completely above the $x$-axis.

Solution:

For a quadratic $ax^2 + bx + c$ to be always positive (curve above $x$-axis):

1. Leading coefficient $a > 0$: $m - 3 > 0 \implies m > 3$
2. Discriminant $< 0$ (no real roots, so curve doesn't cross $x$-axis):

$$\begin{aligned} b^2 - 4ac &< 0 \\ m^2 - 4(m - 3)(m) &< 0 \\ m^2 - 4m^2 + 12m &< 0 \\ -3m^2 + 12m &< 0 \\ 3m^2 - 12m &> 0 \quad \text{(flip when multiplying by } -1) \\ 3m(m - 4) &> 0 \end{aligned}$$

Critical values: $m = 0$ or $m = 4$. Outside region: $m < 0$ or $m > 4$.

Combine both conditions: $m > 3$ AND ($m < 0$ or $m > 4$) → $\boxed{m > 4}$

⚠️ Always need BOTH conditions — discriminant alone isn't enough; the parabola also needs to open the right way.

Question: The expression $-x^2 + 6x + p - 14$ is always negative for all real $x$. Find the range of values of $p$.

Solution:

Leading coefficient is $-1 < 0$, so the parabola opens downward (good — already satisfies the orientation requirement). For "always negative," we additionally need no real roots (curve doesn't touch $x$-axis):

$$\begin{aligned} b^2 - 4ac &< 0 \\ 6^2 - 4(-1)(p - 14) &< 0 \\ 36 + 4p - 56 &< 0 \\ 4p &< 20 \\ p &< 5 \end{aligned}$$ $$\boxed{p < 5}$$

Pair table — quadratic always SIGN means:

Always positive	Always negative
$a > 0$ AND $\Delta < 0$	$a < 0$ AND $\Delta < 0$

Question: A rectangular field has length $(3x - 5)$ m and breadth $(x - 1)$ m. The area is at most $200$ m². Find the range of values of $x$ (assume $x$ gives positive dimensions).

Solution:

Area condition:

$$\begin{aligned} (3x - 5)(x - 1) &\leq 200 \\ 3x^2 - 3x - 5x + 5 &\leq 200 \\ 3x^2 - 8x - 195 &\leq 0 \end{aligned}$$

Solve by quadratic formula:

$$x = \frac{8 \pm \sqrt{64 + 4(3)(195)}}{6} = \frac{8 \pm \sqrt{2404}}{6}$$

$\sqrt{2404} \approx 49.03$, so $x \approx \tfrac{8 + 49.03}{6} \approx 9.50$ or $x \approx \tfrac{8 - 49.03}{6} \approx -6.84$.

Sign of $3x^2 - 8x - 195$: since leading coeff is positive, "$\leq 0$" means BETWEEN roots.

$$-6.84 \leq x \leq 9.50$$

Apply physical constraints (positive dimensions): $3x - 5 > 0 \implies x > \tfrac{5}{3}$ AND $x - 1 > 0 \implies x > 1$. So $x > \tfrac{5}{3}$.

Combined:

$$\boxed{\tfrac{5}{3} < x \leq 9.50}$$

⚠️ Don't forget physical constraints — algebra alone gives a wider range that includes negative widths.

Question: Solve $x^2 - x - 6 < 0$.

Solution:

Step 1. Factorise the LHS:

$$x^2 - x - 6 = (x - 3)(x + 2)$$

Step 2. Find the critical values where the expression equals zero:

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

Step 3. Sign chart — for "$< 0$", we want the expression to be negative. Since the parabola opens upward ($a = 1 > 0$), it is negative BETWEEN the roots.

$$\boxed{-2 < x < 3}$$

Visual: sketch $y = (x-3)(x+2)$ — a smiley parabola crossing the $x$-axis at $-2$ and $3$. The portion below the $x$-axis (negative $y$) is the interval $-2 < x < 3$.

Question: Solve $(x - 3)^2 < x + 27$.

Solution:

Expand and rearrange to the form $\text{(quadratic)} < 0$:

$$\begin{aligned} x^2 - 6x + 9 &< x + 27 \\ x^2 - 7x - 18 &< 0 \\ (x - 9)(x + 2) &< 0 \end{aligned}$$

Critical values: $x = 9$ or $x = -2$. Since the leading coefficient is positive, the parabola is negative between the roots.

$$\boxed{-2 < x < 9}$$

⚠️ Always rearrange to standard form first — comparing the LHS sign chart against zero is the only correct way.

Question: Find the integer values of $x$ satisfying $2x^2 - 9x - 5 \geq 0$.

Solution:

Factorise:

$$2x^2 - 9x - 5 = (2x + 1)(x - 5)$$

Critical values: $x = -\tfrac{1}{2}$ or $x = 5$.

For "$\geq 0$" with positive leading coefficient, we want the OUTSIDE region (including the endpoints):

$$x \leq -\tfrac{1}{2} \quad \text{or} \quad x \geq 5$$

Integer values:

$$\boxed{x \in \{\ldots, -3, -2, -1\} \cup \{5, 6, 7, \ldots\}}$$

Question: Find the values of $h$ for which the line $y + hx = 24$ does not meet the curve $5y + x^2 = 20$.

Solution:

Express both as $y = \ldots$:

$$y = 24 - hx, \qquad y = \frac{20 - x^2}{5} = 4 - \tfrac{1}{5}x^2$$

Equate:

$$\begin{aligned} 24 - hx &= 4 - \tfrac{1}{5}x^2 \\ \tfrac{1}{5}x^2 - hx + 20 &= 0 \\ x^2 - 5hx + 100 &= 0 \end{aligned}$$

"No intersection" → discriminant $< 0$:

$$\begin{aligned} (-5h)^2 - 4(1)(100) &< 0 \\ 25h^2 - 400 &< 0 \\ h^2 &< 16 \\ -4 &< h < 4 \end{aligned}$$ $$\boxed{-4 < h < 4}$$

Question: Solve $\dfrac{x^2 - 4x + 3}{x^2 + 1} > 0$.

Solution:

Step 1. Note $x^2 + 1 > 0$ for all real $x$ (always positive — the discriminant of $x^2 + 1$ is $-4 < 0$, and leading coeff is positive).

Step 2. Since the denominator is always positive, the sign of the fraction matches the sign of the numerator:

$$\text{Solve } x^2 - 4x + 3 > 0$$

Step 3. Factorise:

$$(x - 1)(x - 3) > 0$$

Critical values: $x = 1$ or $x = 3$. Outside region (strict):

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

Key trick: always show the denominator's sign first. If always positive (or always negative) you can drop it and just work with the numerator.

Question: Solve $-3 \leq 2x - 5 < 7$.

Solution:

Split into two simultaneous inequalities and solve each:

$$\begin{aligned} -3 &\leq 2x - 5 \\ 2 &\leq 2x \\ 1 &\leq x \end{aligned} \qquad \text{AND} \qquad \begin{aligned} 2x - 5 &< 7 \\ 2x &< 12 \\ x &< 6 \end{aligned}$$

Both must hold simultaneously:

$$\boxed{1 \leq x < 6}$$

Note: don't forget to keep track of $\leq$ vs $<$ — the endpoint inclusion comes from each inequality separately.

Question: The inequality $2x^2 > 20 - ax$ has solution $x < -5$ or $x > b$. Find $a$ and $b$.

Solution:

Rearrange to standard form:

$$2x^2 + ax - 20 > 0$$

The solution set is the OUTSIDE region, so the parabola opens upward (which it does — leading coeff $2 > 0$ ✓) and the roots are $x = -5$ and $x = b$.

Use sum and product of roots for $2x^2 + ax - 20 = 0$ (which is the boundary equation):

$$\begin{aligned} \text{Sum: } -5 + b &= -\tfrac{a}{2} \\ \text{Product: } -5b &= \tfrac{-20}{2} = -10 \end{aligned}$$

From product: $-5b = -10 \implies b = 2$.

Substitute into sum: $-5 + 2 = -\tfrac{a}{2} \implies -3 = -\tfrac{a}{2} \implies a = 6$.

$$\boxed{a = 6, \quad b = 2}$$

Question: The line $y = mx + 4$ is a tangent to the curve $y = x^2 + 2x + 5$. Find the value(s) of $m$.

Solution:

Step 1. Equate to find points of intersection:

$$mx + 4 = x^2 + 2x + 5$$ $$x^2 + (2 - m)x + 1 = 0$$

Step 2. "Tangent" means exactly one point of contact → discriminant $= 0$:

$$\begin{aligned} (2 - m)^2 - 4(1)(1) &= 0 \\ (2 - m)^2 &= 4 \\ 2 - m &= \pm 2 \\ m &= 0 \text{ or } m = 4 \end{aligned}$$ $$\boxed{m = 0 \text{ or } m = 4}$$

Two values because there are two tangent lines from a point with the same $y$-intercept.