International Baccalaureate · MYP 5

Quadratic Equations

⌚ ~59 min ▢ 7 blocks ✓ 33 exercises ➣ Prerequisites : Expansion and Factorisation, Quadratic Equations, Simultaneous Equations, Radicals and Surds

I Completing the Square when $a \neq 1$

Proposition — Completing the Square

Any quadratic expression of the form $x^2 + bx + c$ (leading coefficient $1$) can be put into the form $(x + p)^2 + q$ by completing the square: $$x^2 + bx + c = \left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c.$$ For a general quadratic $ax^2 + bx + c$ with $a\neq 0$, we first take out the factor $a$: $$ax^2 + bx + c = a\left(x^2 + \frac{b}{a}x + \frac{c}{a}\right)$$ and then complete the square inside the brackets.

Method — Solving by Completing the Square

Given an equation $ax^2+bx+c=0$ with $a\neq 0$:

Step 1: Complete the square to rewrite the left-hand side as $(x + p)^2$ plus or minus a constant, then the equation in the form $(x + p)^2 = q$ (when $a \neq 1$, first divide both sides by $a$).
Step 2: Conclude from the sign of $q$. If $q > 0$: $x + p = \sqrt{q}$ or $x + p = -\sqrt{q}$ (or, in the same way, factorise the difference of squares $(x + p)^2 - \left(\sqrt{q}\right)^2 = 0$ and apply the Null Factor Law), so there are two solutions. If $q = 0$: $x + p = 0$, so there is one solution, $x = -p$. If $q < 0$: there is no solution, since a square is never negative.
Step 3: Solve the linear equations obtained, and check the solutions in the original equation.

Example

Solve each equation by completing the square. Give the exact solutions.

$2x^2 - 8x + 6 = 0$
$3x^2 + 6x - 2 = 0$

Answer

The leading coefficient is not $1$. Since $a \neq 0$, first divide both sides by $a$: the equation keeps the same solutions and its leading coefficient becomes $1$.

Here $a = 2$, and $x^2 - 4x = (x - 2)^2 - 4$: $$\begin{aligned} 2x^2 - 8x + 6 &= 0\\ x^2 - 4x + 3 &= 0 &&\color{gray}\text{(Divide both sides by $2$)}\\ (x - 2)^2 - 4 + 3 &= 0 &&\color{gray}\text{(Complete the square)}\\ (x - 2)^2 - 1^2 &= 0 &&\color{gray}\text{(Simplify)}\\ (x - 2 - 1)(x - 2 + 1) &= 0 &&\color{gray}\text{(Difference of squares)}\\ (x - 3)(x - 1) &= 0 &&\color{gray}\text{(Simplify)}\\ x - 3 = 0 &\text{ or } x - 1 = 0 &&\color{gray}\text{(Null factor law)}\\ x = 3 &\text{ or } x = 1 &&\color{gray}\text{(Solve)}. \end{aligned}$$ Check: $2 \times 3^2 - 8 \times 3 + 6 = 18 - 24 + 6 = 0$ and $2 \times 1^2 - 8 \times 1 + 6 = 2 - 8 + 6 = 0$. The solutions are $1$ and $3$.
Here $a = 3$, and $x^2 + 2x = (x + 1)^2 - 1$: $$\begin{aligned} 3x^2 + 6x - 2 &= 0\\ x^2 + 2x - \frac{2}{3} &= 0 &&\color{gray}\text{(Divide both sides by $3$)}\\ (x + 1)^2 - 1 - \frac{2}{3} &= 0 &&\color{gray}\text{(Complete the square)}\\ (x + 1)^2 &= \frac{5}{3} &&\color{gray}\text{(Add $\tfrac{5}{3}$ to both sides)}\\ x + 1 = \sqrt{\frac{5}{3}} &\text{ or } x + 1 = -\sqrt{\frac{5}{3}} &&\color{gray}\text{(A square equal to a positive number)}\\ x = -1 + \frac{\sqrt{15}}{3} &\text{ or } x = -1 - \frac{\sqrt{15}}{3} &&\color{gray}\text{(Solve, with $\sqrt{\tfrac{5}{3}} = \tfrac{\sqrt{5}}{\sqrt{3}} = \tfrac{\sqrt{15}}{3}$)}. \end{aligned}$$ The solutions are $-1 - \dfrac{\sqrt{15}}{3}$ and $-1 + \dfrac{\sqrt{15}}{3}$ (about $-2.29$ and $0.291$).

Dividing by $a$ brings the equation back to the case $a = 1$ of the Proposition “Completing the Square”; the price is that fractions may appear, as in part 2.

Skills to practice

Solving by Completing the Square when $a \neq 1$

II Solving with the Quadratic Formula

Applying the method of completing the square to the general equation $ax^2 + bx + c = 0$ gives a formula that solves any quadratic equation. This is the quadratic formula.

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Proposition — Quadratic formula

For any quadratic equation $ax^2 + bx + c = 0$, the discriminant, denoted $\Delta$, is defined as $$\Delta = b^2 - 4ac.$$ Its sign determines the number of real solutions:

If $\Delta > 0$, there are two real roots:$$x = \frac{-b - \sqrt{\Delta}}{2a}\text{ or }x = \frac{-b + \sqrt{\Delta}}{2a}$$
If $\Delta = 0$, there is one real root (a double root):$$x = \frac{-b}{2a}.$$
If $\Delta < 0$, there are no real roots.

Proof

Suppose $ax^2 + bx + c = 0$, where $a \neq 0$. $$\begin{aligned} ax^2 + bx + c &= 0 \\ x^2 + \frac{b}{a}x + \frac{c}{a} &= 0 \quad \color{gray}\text{(Divide each term by $a$, since $a \neq 0$)} \\ x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 + \frac{c}{a} &= 0 \quad \color{gray}\text{(Complete the square)} \\ \left(x + \frac{b}{2a}\right)^2 - \frac{b^2 - 4ac}{4a^2} &= 0 \quad \color{gray}\text{(Simplify)} \\ \left(x + \frac{b}{2a}\right)^2 - \frac{\Delta}{4a^2} &= 0 \quad \color{gray}\text{(Where $\Delta = b^2 - 4ac$)}. \end{aligned}$$ Now, consider the cases based on the discriminant $\Delta$:

Case $\Delta \geq 0$: Since $\dfrac{\Delta}{4a^2} \geq 0$, a real square root exists. $$\begin{aligned} \left(x + \frac{b}{2a}\right)^2 &= \frac{\Delta}{4a^2} \\ \left(x + \frac{b}{2a}\right)^2 - \left(\sqrt{\frac{\Delta}{4a^2}}\right)^2 &= 0 \\ \left(x + \frac{b}{2a} - \sqrt{\frac{\Delta}{4a^2}}\right) \left(x + \frac{b}{2a} + \sqrt{\frac{\Delta}{4a^2}}\right) &= 0 \quad \color{gray}\text{(Difference of squares)}. \end{aligned}$$ Applying the Null Factor Law: $$x + \frac{b}{2a} - \sqrt{\frac{\Delta}{4a^2}} = 0 \quad \text{or} \quad x + \frac{b}{2a} + \sqrt{\frac{\Delta}{4a^2}} = 0.$$ Solving these linear equations: $$x = -\frac{b}{2a} + \sqrt{\frac{\Delta}{4a^2}} \quad \text{or} \quad x = -\frac{b}{2a} - \sqrt{\frac{\Delta}{4a^2}}.$$ Since $\sqrt{\dfrac{\Delta}{4a^2}} = \dfrac{\sqrt{\Delta}}{2|a|}$ and the sign $\pm$ covers both cases, this can be written $$x = \frac{-b \pm \sqrt{\Delta}}{2a}.$$
- If $\Delta > 0$, there are two distinct real roots.
- If $\Delta = 0$, there is one real root (double root): $x = -\dfrac{b}{2a}$.
Case $\Delta < 0$: Then $\dfrac{\Delta}{4a^2} < 0$, so $$\left(x + \frac{b}{2a}\right)^2 = \frac{\Delta}{4a^2} < 0.$$ Since the square of a real number is non-negative, there are no real solutions.

Example

Consider the quadratic equation $x^2 + 2x - 3 = 0$.

Find the discriminant.
Hence, state the nature of the roots of the equation.
Solve the equation.

Answer

$x^2 + 2x - 3 = 0$ has $a=1$, $b=2$, $c=-3$.

$\begin{aligned}[t]\Delta &= b^2 - 4ac \\&= (2)^2 - 4(1)(-3) \\&= 4 + 12 \\&= 16\end{aligned}$
As $\Delta > 0$, there are two distinct real solutions.
$\begin{aligned}[t] x &= \frac{-b - \sqrt{\Delta}}{2a} &&\text{ or } && x = \frac{-b + \sqrt{\Delta}}{2a} \\ x &= \frac{-2 - \sqrt{16}}{2 \cdot 1} &&\text{ or } && x = \frac{-2 + \sqrt{16}}{2 \cdot 1} \\ x &= \frac{-2 - 4}{2} &&\text{ or } && x = \frac{-2 + 4}{2} \\ x &= -3 &&\text{ or }&& x = 1 \end{aligned}$

Skills to practice

Solving with the Quadratic Formula Step by Step
Solving Quadratic Equations with the Formula

III Finding Where a Line Meets a Curve

A common problem is to find where the graphs of two functions meet. The intersection points are the points $(x,y)$ that satisfy both equations simultaneously. By setting the function rules equal to each other, we can create a new equation to solve for the x-coordinates of these points.

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Method — Finding Intersection Points

To find the points of intersection between the graphs of two functions $f(x)$ and $g(x)$:

Set the functions equal to each other: $\boldsymbol{f(x) = g(x)}$.
Solve this equation for $x$. The solutions are the $x$-coordinates of the intersection points.
Substitute each $x$-coordinate back into either of the original functions ($f$ or $g$) to find the corresponding y-coordinate.

Example

Find the coordinates of the intersection points of the parabola $f(x) = x^2 - x - 2$ and the line $g(x) = x+1$.

Answer

We find the intersection points by setting $f(x) = g(x)$ and solving for $x$.$$ \begin{aligned}f(x) &= g(x) \\ x^2 - x - 2 &= x+1 \\ x^2 - 2x - 3 &= 0 \\ (x-3)(x+1) &= 0\end{aligned} $$ This gives two solutions for $x$: $x=3$ and $x=-1$.
Now we find the corresponding $y$-coordinates by substituting these values into the linear function $g(x)=x+1$:

If $x=3$, $y = 3+1 = 4$.
If $x=-1$, $y = -1+1 = 0$.

The graphs intersect at the points $\boldsymbol{(-1, 0)}$ and $\boldsymbol{(3, 4)}$.

Skills to practice

Finding Intersection Points Algebraically

IV Going further

Skills to practice

Choosing a Method and Explaining
Solving Problems with the Quadratic Formula
Investigating Patterns with the Quadratic Formula