Worked examples · Sheet 1

Second-degree equations

Theory used: Quadratic formula Discriminant Incomplete equations Zero product property

Exercise 1

Solve the incomplete equations

2 x^{2} - 6 x = 0 and 3 x^{2} - 12 = 0.

2 x^{2} - 6 x = 0 ⟹ 2 x (x - 3) = 0 ⟹ x = 0 or x = 3

3 x^{2} - 12 = 0 ⟹ x^{2} = 4 ⟹ x = \pm 2

The first equation has no constant term, so the common factor $2 x$ is factored out. By the zero product property, the product $2 x (x - 3)$ is zero when $2 x = 0$ or $x - 3 = 0$ , therefore the solutions are $0$ and $3$ .

The second equation has no first-degree term. Isolating $x^{2}$ gives $x^{2} = 4$ ; since $4 > 0$ , the solutions are the two numbers whose square is $4$ , that is, $2$ and $- 2$ .

Exercise 2

Solve the equation

2 x^{2} + x - 3 = 0.

Δ = 1^{2} - 4 \cdot 2 \cdot (- 3) = 1 + 24 = 25

x = \frac{- 1 \pm 25}{2 \cdot 2} = \frac{- 1 \pm 5}{4}

x_{1} = \frac{- 1 - 5}{4} = - \frac{3}{2}, x_{2} = \frac{- 1 + 5}{4} = 1

The equation is complete, with $a = 2$ , $b = 1$ and $c = - 3$ , so the quadratic formula applies. The discriminant equals $25 > 0$ , therefore there are two distinct real solutions.

The square root of the discriminant is $5$ and the formula gives the two values $\frac{- 1 - 5}{4} = - \frac{3}{2}$ and $\frac{- 1 + 5}{4} = 1$ . Checking by substitution confirms the result: for $x = 1$ we get $2 + 1 - 3 = 0$ .

Exercise 3

Solve the equations

4 x^{2} - 12 x + 9 = 0 and x^{2} + 2 x + 5 = 0.

4 x^{2} - 12 x + 9 = 0 : Δ = 144 - 144 = 0, x = \frac{12}{8} = \frac{3}{2}

x^{2} + 2 x + 5 = 0 : Δ = 4 - 20 = - 16 < 0

The first equation has the double solution $x = \frac{3}{2}$ ; the second has no real solutions.

In the first equation the discriminant is zero, so there is only one real solution, a double one, given by $x = - \frac{b}{2 a} = \frac{12}{8} = \frac{3}{2}$ . Indeed, the trinomial is the square of a binomial, since $4 x^{2} - 12 x + 9 = (2 x - 3)^{2}$ , and a square vanishes only where its base vanishes.

In the second equation the discriminant is negative, therefore the quadratic formula would require the square root of a negative number, which does not exist in the real numbers. Hence the equation has no real solutions.