Quadratic Equation: Important Questions (Part 1)

This lesson revises the essential facts and formulae for quadratic equations, then works through several equations: four solved by splitting the middle term and one by the quadratic formula.

Key facts about a quadratic equation

The general (standard) form of a quadratic equation is

$ax^2 + bx + c = 0, \qquad a \neq 0,$

where $a$, $b$, $c$ are real numbers and $b$ or $c$ may be $0$. Its degree is $2$, and a quadratic equation has at most $2$ roots (it may have $2$, $1$, or no real roots).

The quadratic formula and the discriminant

For $ax^2 + bx + c = 0$ with $a \neq 0$, the roots are given by

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

where $a$ is the coefficient of $x^2$, $b$ the coefficient of $x$, and $c$ the constant term. The discriminant is

$D = b^2 - 4ac.$

Nature of the roots

The sign of the discriminant decides the nature of the roots.

• If $D < 0$: no real roots.
• If $D = 0$: two equal real roots (one repeated root).
• If $D > 0$: two distinct real roots.

So $D \ge 0$ gives real roots.

Handy formulae for word problems

These set-up formulae are useful when a word problem leads to a quadratic.

Two-digit numbers

If the tens digit is $x$ and the units digit is $y$, then the number is

$10x + y,$

and with the digits reversed it becomes

$10y + x.$

Distance, speed, and time

$\text{distance} = \text{speed} \times \text{time}, \qquad \text{time} = \frac{\text{distance}}{\text{speed}}, \qquad \text{speed} = \frac{\text{distance}}{\text{time}}.$

Unit conversions

$\text{km/h} \to \text{m/s}: \ \times \tfrac{5}{18}, \qquad \text{m/s} \to \text{km/h}: \ \times \tfrac{18}{5}.$

Also: hours to minutes, multiply by $60$; minutes to hours, divide by $60$; hours to seconds, multiply by $60 \times 60$; minutes to seconds, multiply by $60$.

Solving by splitting the middle term

To factorise, find two numbers whose product is $a \times c$ and whose sum is $b$, then split the middle term.

Example 1: $x^2 + 8x + 15 = 0$

Product $= 15$, sum $= 8$, so the numbers are $5$ and $3$. Since the coefficient of $x^2$ is $1$, write directly:

$(x + 5)(x + 3) = 0 \implies x = -5 \ \text{or}\ x = -3.$

Solution: $x = -5, \ -3$.

Example 2: $6x^2 + 11x - 10 = 0$

Product $= 6 \times (-10) = -60$, sum $= 11$. The numbers are $15$ and $-4$ (their product is $-60$, their sum is $11$). Split:

$6x^2 + 15x - 4x - 10 = 0$ $3x(2x + 5) - 2(2x + 5) = 0$ $(2x + 5)(3x - 2) = 0 \implies x = -\tfrac{5}{2} \ \text{or}\ x = \tfrac{2}{3}.$

Solution: $x = -\tfrac{5}{2}, \ \tfrac{2}{3}$.

Example 3: $6x^2 - 19x + 10 = 0$

Product $= 60$, sum $= -19$, so both numbers are negative: $-15$ and $-4$. Split:

$6x^2 - 15x - 4x + 10 = 0$ $3x(2x - 5) - 2(2x - 5) = 0$ $(2x - 5)(3x - 2) = 0 \implies x = \tfrac{5}{2} \ \text{or}\ x = \tfrac{2}{3}.$

Solution: $x = \tfrac{5}{2}, \ \tfrac{2}{3}$.

Example 4: $6x^2 - 11x - 10 = 0$

Product $= -60$, sum $= -11$. The numbers are $-15$ and $4$. Split:

$6x^2 - 15x + 4x - 10 = 0$ $3x(2x - 5) + 2(2x - 5) = 0$ $(2x - 5)(3x + 2) = 0 \implies x = \tfrac{5}{2} \ \text{or}\ x = -\tfrac{2}{3}.$

Solution: $x = \tfrac{5}{2}, \ -\tfrac{2}{3}$.

Solving by the quadratic formula

Example: $x^2 + 15x + 36 = 0$

First check it is in standard form (it is), so $a = 1$, $b = 15$, $c = 36$. Find the discriminant:

$D = b^2 - 4ac = 15^2 - 4(1)(36) = 225 - 144 = 81 > 0,$

so there are two distinct real roots, and $\sqrt{D} = \sqrt{81} = 9$. Apply the formula:

$x = \frac{-15 \pm 9}{2(1)} = \frac{-15 + 9}{2}, \ \frac{-15 - 9}{2} = \frac{-6}{2}, \ \frac{-24}{2} = -3, \ -12.$

Solution: $x = -3, \ -12$.

Key takeaways

• A quadratic equation $ax^2 + bx + c = 0$ ($a \neq 0$) has degree $2$ and at most two roots.
• The discriminant $D = b^2 - 4ac$ decides the nature of the roots: $D < 0$ none, $D = 0$ equal, $D > 0$ distinct.
• To split the middle term, find two numbers with product $a \times c$ and sum $b$; otherwise use $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.