Here we go beyond quadratics to cubic and quartic equations, and to simultaneous equations in two unknowns.
The real roots of a cubic equation are the x-coordinates where its graph meets the x-axis, so once the factor theorem gives one root, synthetic division lowers the degree and the remaining quadratic can be solved.
The solutions of simultaneous equations are the points where two graphs meet, and a pair made of a linear and a quadratic equation is solved by substituting one into the other to get an equation in one unknown.
In the first picture the number of points where the cubic meets the x-axis changes with p and q; in the second, the intersections of the parabola and the line change with the gradient k.
Finding the roots of a cubic from its graph
The real roots of the cubic equation x³ + px + q = 0 are the x-coordinates of the points where the graph meets the x-axis. Changing the coefficients changes the number of real roots between 1, 2 and 3.
-3
2
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The graph of y = x³ + px + q with the points where it meets the x-axis and the number of distinct real roots
💡 Real roots of a cubic
①A cubic graph runs from the bottom left to the top right, so it meets the x-axis at least once; a cubic equation always has at least one real root
②When p is negative and large in size, the graph bends enough to cross the x-axis three times
③Where the graph touches the x-axis there is a repeated root: the starting example x³ − 3x + 2 = (x − 1)²(x + 2) touches at x = 1
④To solve, find one root with the factor theorem and reduce to a quadratic with synthetic division
Solving cubics and quartics
Using the factor theorem
integer root candidates → synthetic division → reduce to a quadratic
Substitute divisors of the constant term to find a value that gives 0
Using substitution
x⁴ + ax² + b = 0 → substitute x² = t
The quartic becomes a quadratic in t
🔑 Order of steps
①First look for a common factor
②Substitute divisors of the constant term to find one root
③Lower the degree by one with synthetic division
④Solve the quadratic that is left by factorising or with the formula
Simultaneous equations and intersections
The solutions of simultaneous equations are the coordinates of the points where the two graphs meet. A linear and a quadratic equation together are solved by substituting one into the other.
2
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The two points where the parabola y = x² and the line y = kx + 1 meet
📐 Solving simultaneous equations
①Substitute one equation into the other to get an equation in one unknown
②x² = kx + 1 rearranges to x² − kx − 1 = 0
③Its discriminant is D = k² + 4, positive for every k, so the line in the picture always meets the parabola at two points
④The coordinates of the two intersections are the solutions
Special equations
Equations with absolute values
|f(x)| = g(x) → f(x) = g(x) or f(x) = −g(x)
Split into two cases, checking that g(x) ≥ 0
Irrational equations
√(f(x)) = g(x) → f(x) = g(x)², g(x) ≥ 0
After squaring, always check in the original equation
Worked examples and unit test
Example 1
Find the real roots of x³ − 8 = 0.
1
Factorise using the difference of cubes.
x³ − 2³ = (x − 2)(x² + 2x + 4) = 0
2
x − 2 = 0 gives the real root; x² + 2x + 4 = 0 has discriminant 4 − 16 < 0, so its roots are complex.
x = 2 (the roots of x² + 2x + 4 = 0 are complex)
▸ x = 2
x³ − 8 = 0 has three roots in all, but only one of them is real, and the graph y = x³ − 8 meets the x-axis only once.
Example 2
Solve the simultaneous equations x + y = 5 and x² + y² = 13.
1
From x + y = 5, y = 5 − x; substitute into the second equation.
x² + (5 − x)² = 13
2
This simplifies to 2x² − 10x + 12 = 0, that is x² − 5x + 6 = 0, so (x − 2)(x − 3) = 0.
x² − 5x + 6 = 0 ⇒ (x, y) = (2, 3) or (3, 2)
▸ (x, y) = (2, 3) or (3, 2)
It also means the circle x² + y² = 13 and the line x + y = 5 meet at the two points (2, 3) and (3, 2).
Summary
Key to higher-degree equations
factor theorem + synthetic division → lower the degree
Break cubics and quartics down to quadratics or lower
Grade-10 education-office assessment type
Find the sum of the three real roots of x³ − 6x² + 11x − 6 = 0.
①3
②4
③5
④6
⑤7
▸ ④ 6
1
For ax³ + bx² + cx + d = 0, the sum of the roots is −b/a.
α + β + γ = −b/a
2
Substitute a = 1, b = −6.
α + β + γ = −(−6)/1 = 6
🎯 Points that often come up
①A cubic equation always has at least one real root
②For a quartic with only even powers such as x² and x⁴, try x² = t
③Turn simultaneous equations into one equation in one unknown by substitution
④For irrational equations, square both sides and weed out extraneous roots
⑤For equations with absolute values, split into cases and check the condition in each