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CIE A-Level Maths Study Notes

1.1.6 Equations Reducible to Quadratics

Understanding equations that can be reduced to quadratic form is essential. These equations, often appearing complex at first glance, can be transformed into a quadratic equation through substitution. This topic is particularly relevant for problems involving higher powers and trigonometric functions.

Recognising Equations Reducible to Quadratics

Identifying an equation that can be transformed into a quadratic form involves looking for patterns akin to the standard quadratic equation ax2+bx+c=0ax^2 + bx + c = 0. These patterns may involve variables raised to different powers or variables within trigonometric functions.

Key Indicators:

  • Terms that can be squared or involve square roots.
  • Equations with trigonometric functions where the variable appears in different forms, like sin(x)\sin(x) and sin2(x)\sin^2(x).
  • Higher power terms where one term is a power of another, such as x4x^4and x2x^2.

Methods of Substitution

Substitution simplifies these equations into a quadratic form. The process involves replacing a complex part of the equation with a simpler variable, typically uu, to transform it into a quadratic equation.

Steps for Substitution:

1. Identify the Substitution: Choose a part of the equation that, when substituted, will simplify it to a quadratic form.

2. Rewrite the Equation: Replace the identified part with uu and rewrite the equation.

3. Solve the Quadratic Equation: Solve the new equation as a standard quadratic.

4. Back-Substitute: Replace uu with its original expression and solve for xx.

Examples:

Example 1: Higher Power Equation

Solve x45x2+6=0x^4 - 5x^2 + 6 = 0

Solution:

1. Substitute: Let u=x2u = x^2. The equation becomes u25u+6=0u^2 - 5u + 6 = 0.

2. Factor: (u2)(u3)=0(u - 2)(u - 3) = 0, so u=2u = 2 or u=3u = 3.

3. Reverse: Put x2x^2 back for uu, getting x2=2x^2 = 2 and x2=3x^2 = 3.

4. Solve: For x2=2x^2 = 2, x=±2x = \pm\sqrt{2}. For x2=3x^2 = 3, x=±3x = \pm\sqrt{3}.

Answers: x=±2x = \pm\sqrt{2}, x=±3x = \pm\sqrt{3}.

Example 2: Trigonometric Function

Solve sin2(x)3sin(x)+2=0.\sin^2(x) - 3\sin(x) + 2 = 0.

Solution:

1. Substitute: Let u=sin(x)u = \sin(x). Equation is u23u+2=0u^2 - 3u + 2 = 0.

2. Factor: (u1)(u2)=0(u - 1)(u - 2) = 0, thus u=1u = 1 or u=2u = 2.

3. Reject: u=2u = 2 is invalid as sin(x)\sin(x) is between -1 and 1.

4. Reverse: For u=1u = 1, sin(x)=1\sin(x) = 1.

5. Solve: x=π2+2kπx = \frac{\pi}{2} + 2k\pi, where kk is an integer.

Answer: x=π2+2kπx = \frac{\pi}{2} + 2k\pi

Advanced Techniques

Working with Higher Powers

In equations with higher powers, such as x62x3+1=0x^6 - 2x^3 + 1 = 0, the substitution method still applies. Here, setting u=x3u = x^3 transforms the equation into a quadratic form. Solving the quadratic equation and then back-substituting uu with x3x^3 leads to the original variable solutions.

Trigonometric Transformations

For equations involving trigonometric functions, like cos2(x)4cos(x)+3=0\cos^2(x) - 4\cos(x) + 3 = 0, the substitution u=cos(x)u = \cos(x) can be used.

After solving the quadratic in uu, the solutions in terms of xx can be found using trigonometric identities and inverse functions.

Examples:

Example 1: Trigonometric Equation

Solve 4cos2(x)3cos(x)1=04\cos^2(x) - 3\cos(x) - 1 = 0 for xx.

Solution:

1. Let u=cos(x)u = \cos(x), so 4u23u1=04u^2 - 3u - 1 = 0.

2. Apply quadratic formula: u=3±258u = \frac{3 \pm \sqrt{25}}{8}.

3. Simplify to find u=1u = 1 or u=14u = -\frac{1}{4}.

4. For u=1u = 1, x=2πnx = 2\pi n (since cos(x)=1\cos(x) = 1 at these points).

5. For u=14u = -\frac{1}{4}, x=arccos(14)+2πnx = \arccos(-\frac{1}{4}) + 2\pi n.

Answers: x=2πnx = 2\pi n and x=arccos(14)+2πnx = \arccos(-\frac{1}{4}) + 2\pi n, for any integer nn.

Example 2: Higher Power Equation

Solve x65x3+6=0x^6 - 5x^3 + 6 = 0 for xx.

Solution:

1. Let u=x3u = x^3, changing the equation to u25u+6=0u^2 - 5u + 6 = 0.

2. Factorise to (u2)(u3)=0(u - 2)(u - 3) = 0, giving u=2u = 2 or u=3u = 3.

3. Replace uu with x3x^3 to get x3=2x^3 = 2 and x3=3x^3 = 3.

4. Solve each to find x=23x = \sqrt[3]{2} and x=33x = \sqrt[3]{3}.

Answers: x=23x = \sqrt[3]{2} and x=33x = \sqrt[3]{3}.

Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
LinkedIn
Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.

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