What’s New in Grade 11?#
In Grade 10, you solved linear equations. Now we tackle equations where $x$ appears as $x^2$ — these have two solutions (or sometimes one, or none).
1. Solving by Factoring#
Get everything to one side (= 0), factor, then use the Zero Product Rule: if $ab = 0$, then $a = 0$ or $b = 0$.
Worked Example#
$x^2 - 5x + 6 = 0$
$(x - 2)(x - 3) = 0$
$x = 2$ or $x = 3$
Another Example#
$2x^2 + x = 6$
$2x^2 + x - 6 = 0$
$(2x - 3)(x + 2) = 0$
$x = \frac{3}{2}$ or $x = -2$
⚠️ NEVER divide both sides by $x$ — you’ll lose the solution $x = 0$.
$x^2 = 5x$ → WRONG: $x = 5$. CORRECT: $x^2 - 5x = 0 \Rightarrow x(x-5) = 0 \Rightarrow x = 0$ or $x = 5$.
2. The Quadratic Formula#
When you can’t factorise, use:
$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$For $ax^2 + bx + c = 0$.
Worked Example#
$2x^2 + 3x - 7 = 0$
$a = 2$, $b = 3$, $c = -7$
$x = \frac{-3 \pm \sqrt{9 + 56}}{4} = \frac{-3 \pm \sqrt{65}}{4}$
$x = \frac{-3 + 8.062}{4} = 1.27$ or $x = \frac{-3 - 8.062}{4} = -2.77$
3. The Discriminant ($\Delta$)#
The expression under the square root tells you EVERYTHING about the solutions before you even solve:
$$ \Delta = b^2 - 4ac $$| Value of $\Delta$ | Nature of Roots |
|---|---|
| $\Delta > 0$, perfect square | Two real, rational, unequal roots |
| $\Delta > 0$, not perfect square | Two real, irrational, unequal roots |
| $\Delta = 0$ | Two equal (repeated) real roots |
| $\Delta < 0$ | No real roots (non-real) |
Why it works#
The formula has $\sqrt{\Delta}$ in it. If $\Delta < 0$, you’re taking the square root of a negative — impossible in real numbers. If $\Delta = 0$, the $\pm$ gives the same answer both times.
Worked Example: Finding $k$#
For which values of $k$ will $x^2 + kx + 9 = 0$ have equal roots?
Equal roots means $\Delta = 0$:
$k^2 - 4(1)(9) = 0$
$k^2 = 36$
$k = \pm 6$
Worked Example: Real roots#
For which values of $p$ will $2x^2 - 4x + p = 0$ have real roots?
Real roots means $\Delta \geq 0$:
$16 - 8p \geq 0$
$p \leq 2$
4. Quadratic Inequalities#
The Method#
- Solve the corresponding equation ($= 0$) to find the critical values.
- Sketch a mini-parabola.
- Read the answer from the sketch.
Worked Example 1: $x^2 - 4x - 5 \leq 0$#
Step 1: $(x - 5)(x + 1) = 0 \Rightarrow x = 5$ or $x = -1$
Step 2: The parabola opens upward ($a > 0$), so it’s BELOW zero BETWEEN the roots.
Step 3: $-1 \leq x \leq 5$
Worked Example 2: $x^2 - 4x - 5 > 0$#
Same roots, but now we want where the parabola is ABOVE zero — OUTSIDE the roots:
$x < -1$ or $x > 5$
Worked Example 3: $-x^2 + 2x + 3 \geq 0$#
Multiply by $-1$ (flip the sign!): $x^2 - 2x - 3 \leq 0$
$(x - 3)(x + 1) = 0 \Rightarrow x = 3$ or $x = -1$
Upward parabola, below zero between roots: $-1 \leq x \leq 3$
Pro tip: If $a < 0$, multiply through by $-1$ first (and flip the inequality). Then sketch the standard upward parabola.
5. Simultaneous Equations (Linear + Quadratic)#
The Method#
- Make one variable the subject in the linear equation.
- Substitute into the quadratic equation.
- Solve the resulting quadratic.
- Substitute back for the other variable.
Worked Example#
$y = x + 1$ … (1)
$x^2 + y^2 = 13$ … (2)
Substitute (1) into (2):
$x^2 + (x + 1)^2 = 13$
$x^2 + x^2 + 2x + 1 = 13$
$2x^2 + 2x - 12 = 0$
$x^2 + x - 6 = 0$
$(x + 3)(x - 2) = 0$
$x = -3$ or $x = 2$
From (1): $y = -2$ or $y = 3$
Solutions: $(-3; -2)$ and $(2; 3)$
What does this mean graphically?#
These are the intersection points of the line and the circle. A line can intersect a circle at 0, 1, or 2 points. The discriminant of the resulting quadratic tells you which case it is.
6. Equations with Fractions#
Worked Example#
$\frac{3}{x-2} + 1 = \frac{x}{x-2}$
Multiply through by $(x - 2)$:
$3 + (x - 2) = x$
$3 + x - 2 = x$
$1 = 0$???
No solution! This happens when $x = 2$ makes the denominator zero, and there’s no other valid $x$.
Always state restrictions: Before solving, note which values of $x$ make denominators zero. These are EXCLUDED from the answer.
🚨 Common Mistakes#
- Dividing by $x$: NEVER divide both sides by $x$ (or $\sin\theta$). Move everything to one side and factor instead.
- Sign error in the quadratic formula: $b^2 - 4ac$ — watch the signs. If $c = -7$, then $-4ac = -4(2)(-7) = +56$.
- Inequality sign after multiplying by $-1$: FLIP IT. $-x^2 + 3 > 0$ becomes $x^2 - 3 < 0$.
- Simultaneous equations — expanding $(x+1)^2$: It’s $x^2 + 2x + 1$, NOT $x^2 + 1$. The middle term!
- Forgetting restrictions: $\frac{3}{x-2}$ is undefined when $x = 2$. State this upfront.
💡 Pro Tip: The “Parabola Sketch” Shortcut#
For quadratic inequalities, you don’t need a perfect graph. Just:
- Mark the roots on a number line.
- Draw a quick U-shape (if $a > 0$) or ∩-shape (if $a < 0$).
- Shade above or below the x-axis depending on the inequality sign.
🔗 Related Grade 11 topics:
- Surds & Exponential Equations — surd equations become quadratics after squaring
- The Parabola — the x-intercepts of a parabola = the roots of the quadratic equation
- Trig Identities & Equations — trig equations often reduce to quadratics (e.g., $2\cos^2\theta - 1 = 0$)
📌 Grade 10 foundation: Solving Equations & Inequalities — linear equations, literal equations, and simultaneous equations
📌 Grade 12 extension: Algebra, Equations & Inequalities — revision and extension for matric