Worked examples · Sheet 2

Higher-degree equations

Theory used: Solving by factorization Zero product property Biquadratic equations

Exercise 1

Solve the equation

x^{3} - 4 x = 0.

x^{3} - 4 x = x (x^{2} - 4) = x (x - 2) (x + 2) = 0

x = 0, x = 2, x = - 2

All terms contain the factor $x$ , so the first step is factoring out the common factor. The factor $x^{2} - 4$ is a difference of squares, therefore it factors as $(x - 2) (x + 2)$ and the left-hand side becomes a product of three first-degree factors.

By the zero product property, the product is zero when at least one factor is zero, so the solutions are $0$ , $2$ and $- 2$ . The equation has degree $3$ and three solutions were found, therefore they are all of them.

Exercise 2

Solve the equation

x^{3} + 3 x^{2} - 4 = 0.

Setting $P (x) = x^{3} + 3 x^{2} - 4$ , we have $P (1) = 1 + 3 - 4 = 0$ .

111314044 - 4 40

(x - 1) (x^{2} + 4 x + 4) = (x - 1) (x + 2)^{2} = 0

x = 1, x = - 2 (double)

The polynomial has leading coefficient $1$ , so any rational roots are integers and divide the constant term $- 4$ . The candidate $1$ is a root, since $P (1) = 0$ , therefore by Ruffini’s theorem $x - 1$ divides $P$ . In Ruffini’s scheme the term of degree $1$ is missing, so its coefficient is $0$ .

The quotient $x^{2} + 4 x + 4$ contains the squares of $x$ and of $2$ and twice their product $4 x$ , so it is the square of the binomial $x + 2$ . By the zero product property the solutions are $1$ and $- 2$ ; the solution $- 2$ is double, because the factor $x + 2$ appears squared.

Exercise 3

Solve the biquadratic equation

x^{4} + 3 x^{2} - 4 = 0.

With the substitution $t = x^{2}$ :

t^{2} + 3 t - 4 = 0, Δ = 9 + 16 = 25, t = \frac{- 3 \pm 5}{2}

t_{1} = 1, t_{2} = - 4

x^{2} = 1 ⟹ x = \pm 1; x^{2} = - 4 no real solutions

The solutions are $x = 1$ and $x = - 1$ .

The equation contains only the degrees $4$ , $2$ and $0$ , so it is biquadratic and the substitution $t = x^{2}$ turns it into a second-degree equation in $t$ . The discriminant equals $25$ and the solutions are $t_{1} = \frac{- 3 + 5}{2} = 1$ and $t_{2} = \frac{- 3 - 5}{2} = - 4$ .

One returns to the variable $x$ by solving $x^{2} = t$ for each value. From $x^{2} = 1$ the solutions $\pm 1$ follow. The equation $x^{2} = - 4$ has no real solutions, because a square is never negative, so the value $t_{2} = - 4$ produces no solutions and the original equation has exactly the two solutions $\pm 1$ .