IB Mathematics AA SL Solving equations Study Notes
IB Mathematics AA SL Solving equations Study Notes Offer a clear explanation of Solving equations , including various formula, rules, exam style questions as example to explain the topics. Worked Out examples and common problem types provided here will be sufficient to cover for topic Solving equations
Solving Equations
Solving equations involves finding the value(s) of the variable(s) that satisfy the equation. There are different types of equations and methods to solve them. Here, we will cover linear, quadratic, and exponential equations.
1. Solving Linear Equations
A linear equation is an equation of the form \( ax + b = 0 \), where \( a \neq 0 \).
Steps to Solve:
- Isolate the variable \( x \) by moving all terms involving \( x \) to one side of the equation.
- Simplify the equation to find \( x \).
Example: Solve \( 2x – 4 = 10 \)
- Add \( 4 \) to both sides: \( 2x = 14 \)
- Divide both sides by \( 2 \): \( x = 7 \)
2. Solving Quadratic Equations
A quadratic equation is of the form \( ax^2 + bx + c = 0 \), where \( a \neq 0 \).
Methods to Solve:
- Factorization: Express the quadratic equation as \( (x – p)(x – q) = 0 \) and solve for \( x \).
- Quadratic Formula: \( x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \)
- Completing the Square: Rewrite the equation in the form \( (x – h)^2 = k \).
Example: Solve \( x^2 – 5x + 6 = 0 \)
- Factorization: \( (x – 2)(x – 3) = 0 \)
- Solutions: \( x = 2 \) or \( x = 3 \)
3. Solving Exponential Equations
An exponential equation has the form \( a^x = b \), where \( a > 0 \) and \( a \neq 1 \).
Steps to Solve:
- If possible, rewrite both sides with the same base and solve for the exponent \( x \).
- If not, take the logarithm of both sides and use logarithmic properties to solve for \( x \).
Example: Solve \( 2^x = 16 \)
- Rewrite \( 16 \) as \( 2^4 \): \( 2^x = 2^4 \)
- Since the bases are equal, \( x = 4 \)
4. Solving Logarithmic Equations
A logarithmic equation has the form \( \log_a(x) = b \), which is equivalent to \( x = a^b \).
Steps to Solve:
- Rewrite the equation in its exponential form.
- Solve the resulting exponential equation.
Example: Solve \( \log_2(x) = 3 \)
- Rewrite as \( x = 2^3 \)
- Solve: \( x = 8 \)
5. Solving Rational Equations
A rational equation involves fractions with polynomials in the numerator and denominator, such as \( \frac{a}{b} = \frac{c}{d} \).
Steps to Solve:
- Cross-multiply to eliminate the fractions.
- Simplify and solve the resulting equation.
Example: Solve \( \frac{x + 1}{2} = \frac{3x}{4} \)
- Cross-multiply: \( 4(x + 1) = 2(3x) \)
- Simplify: \( 4x + 4 = 6x \)
- Solve: \( 2x = 4 \implies x = 2 \)
Examples on Solving Equations
Example 1: Solving a Quadratic Equation Analytically
Consider the equation: \( x^2 – 5x + 6 = 0 \)
To solve this equation, we can use factorization:
- Rewrite the equation: \( (x – 2)(x – 3) = 0 \)
- Set each factor equal to zero: \( x – 2 = 0 \) or \( x – 3 = 0 \)
- Solutions: \( x = 2 \) and \( x = 3 \)
Answer: The roots of the equation are \( x = 2 \) and \( x = 3 \).
Example 2: Solving an Exponential Equation Graphically
Consider the equation: \( e^x = 5 – 2x \)
To solve this equation graphically:
- Plot the functions \( y = e^x \) and \( y = 5 – 2x \) on the same graph.
- Identify the point of intersection.
Answer: The approximate solution is the x-coordinate of the intersection point, say \( x \approx 1.1 \).
Example 3: Solving an Equation Using the Quadratic Formula
Consider the equation: \( 2x^2 – 4x – 6 = 0 \)
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \):
- Here, \( a = 2 \), \( b = -4 \), and \( c = -6 \).
- Calculate the discriminant: \( b^2 – 4ac = (-4)^2 – 4(2)(-6) = 16 + 48 = 64 \).
- Find the roots:
- \( x = \frac{-(-4) \pm \sqrt{64}}{2(2)} = \frac{4 \pm 8}{4} \)
- \( x = \frac{4 + 8}{4} = 3 \) and \( x = \frac{4 – 8}{4} = -1 \)
Answer: The solutions are \( x = 3 \) and \( x = -1 \).
Example 4: Solving a Trigonometric Equation Using Technology
Consider the equation: \( \sin(x) = \frac{1}{2} \)
To solve this equation using a graphing tool:
- Graph the functions \( y = \sin(x) \) and \( y = \frac{1}{2} \).
- Find the points of intersection in the interval \( 0 \leq x < 2\pi \).
Answer: The solutions are \( x = \frac{\pi}{6} \) and \( x = \frac{5\pi}{6} \).
Example 5: Solving a Polynomial Equation Using Factorization
Consider the equation: \( x^3 – 3x^2 – 4x + 12 = 0 \)
To solve this equation by factorization:
- Group the terms: \( (x^3 – 3x^2) – (4x – 12) = 0 \).
- Factor out common terms: \( x^2(x – 3) – 4(x – 3) = 0 \).
- Factor by grouping: \( (x^2 – 4)(x – 3) = 0 \).
- Factor further: \( (x – 2)(x + 2)(x – 3) = 0 \).
- Solutions: \( x = 2 \), \( x = -2 \), and \( x = 3 \).
Answer: The roots of the equation are \( x = 2 \), \( x = -2 \), and \( x = 3 \).
IB Mathematics AA SL Solving equations Exam Style Worked Out Questions
Question
The graph of a polynomial function f of degree 4 is shown below.
Given that \({(x + {\text{i}}y)^2} = – 5 + 12{\text{i}},{\text{ }}x,{\text{ }}y \in \mathbb{R}\) . Show that
(i) \({x^2} – {y^2} = – 5\) ;
(ii) \(xy = 6\) .[2]
Hence find the two square roots of \( – 5 + 12{\text{i}}\) .[5]
For any complex number z , show that \({(z^*)^2} = ({z^2})^*\) .[3]
Hence write down the two square roots of \( – 5 – 12{\text{i}}\) .[2]
Explain why, of the four roots of the equation \(f(x) = 0\) , two are real and two are complex.[2]
The curve passes through the point \(( – 1,\, – 18)\) . Find \(f(x)\) in the form
\(f(x) = (x – a)(x – b)({x^2} + cx + d),{\text{ where }}a,{\text{ }}b,{\text{ }}c,{\text{ }}d \in \mathbb{Z}\) .[5]
Find the two complex roots of the equation \(f(x) = 0\) in Cartesian form.[2]
Draw the four roots on the complex plane (the Argand diagram).[2]
Express each of the four roots of the equation in the form \(r{{\text{e}}^{{\text{i}}\theta }}\) .[6]
▶️Answer/Explanation
Markscheme
(i) \({(x + {\text{i}}y)^2} = – 5 + 12{\text{i}}\)
\({x^2} + 2{\text{i}}xy + {{\text{i}}^2}{y^2} = – 5 + 12{\text{i}}\) A1
(ii) equating real and imaginary parts M1
\({x^2} – {y^2} = – 5\) AG
\(xy = 6\) AG
[2 marks]
substituting M1
EITHER
\({x^2} – \frac{{36}}{{{x^2}}} = – 5\)
\({x^4} + 5{x^2} – 36 = 0\) A1
\({x^2} = 4,\, – 9\) A1
\(x = \pm 2\) and \(y = \pm 3\) (A1)
OR
\(\frac{{36}}{{{y^2}}} – {y^2} = – 5\)
\({y^4} – 5{y^2} – 36 = 0\) A1
\({y^2} = 9,\, – 4\) A1
\({y^2} = \pm 3\) and \(x = \pm 2\) (A1)
Note: Accept solution by inspection if completely correct.
THEN
the square roots are \((2 + 3{\text{i}})\) and \(( – 2 – 3{\text{i}})\) A1
[5 marks]
EITHER
consider \(z = x + {\text{i}}y\)
\(z^* = x – {\text{i}}y\)
\({(z^*)^2} = {x^2} – {y^2} – 2{\text{i}}xy\) A1
\(({z^2}) = {x^2} – {y^2} + 2{\text{i}}xy\) A1
\(({z^2})^* = {x^2} – {y^2} – 2{\text{i}}xy\) A1
\({(z^*)^2} = ({z^2})^*\) AG
OR
\(z^* = r{{\text{e}}^{ – {\text{i}}\theta }}\)
\({(z^*)^2} = {r^2}{{\text{e}}^{ – 2{\text{i}}\theta }}\) A1
\({z^2} = {r^2}{{\text{e}}^{2{\text{i}}\theta }}\) A1
\(({z^2})^* = {r^2}{{\text{e}}^{ – 2{\text{i}}\theta }}\) A1
\({(z^*)^2} = ({z^2})^*\) AG
[3 marks]
\((2 – 3{\text{i}})\) and \(( – 2 + 3{\text{i}})\) A1A1
[2 marks]
the graph crosses the x-axis twice, indicating two real roots R1
since the quartic equation has four roots and only two are real, the other two roots must be complex R1
[2 marks]
\(f(x) = (x + 4)(x – 2)({x^2} + cx + d)\) A1A1
\(f(0) = – 32 \Rightarrow d = 4\) A1
Since the curve passes through \(( – 1,\, – 18)\),
\( – 18 = 3 \times ( – 3)(5 – c)\) M1
\(c = 3\) A1
Hence \(f(x) = (x + 4)(x – 2)({x^2} + 3x + 4)\)
[5 marks]
\(x = \frac{{ – 3 \pm \sqrt {9 – 16} }}{2}\) (M1)
\( \Rightarrow x = – \frac{3}{2} \pm {\text{i}}\frac{{\sqrt 7 }}{2}\) A1
[2 marks]
A1A1
Note: Accept points or vectors on complex plane.
Award A1 for two real roots and A1 for two complex roots.
[2 marks]
real roots are \(4{{\text{e}}^{{\text{i}}\pi }}\) and \(2{{\text{e}}^{{\text{i}}0}}\) A1A1
considering \( – \frac{3}{2} \pm {\text{i}}\frac{{\sqrt 7 }}{2}\)
\(r = \sqrt {\frac{9}{4} + \frac{7}{4}} = 2\) A1
finding \(\theta \) using \(\arctan \left( {\frac{{\sqrt 7 }}{3}} \right)\) M1
\(\theta = \arctan \left( {\frac{{\sqrt 7 }}{3}} \right) + \pi {\text{ or }}\theta = \arctan \left( { – \frac{{\sqrt 7 }}{3}} \right) + \pi \) A1
\( \Rightarrow z = 2{{\text{e}}^{{\text{i}}\left( {\arctan \left( {\frac{{\sqrt 7 }}{3}} \right) + \pi } \right)}}{\text{ or}} \Rightarrow z = 2{{\text{e}}^{{\text{i}}\left( {\arctan \left( {\frac{{ – \sqrt 7 }}{3}} \right) + \pi } \right)}}\) A1
Note: Accept arguments in the range \( – \pi {\text{ to }}\pi {\text{ or }}0{\text{ to }}2\pi \) .
Accept answers in degrees.
[6 marks]
Question
The quadratic equation \({x^2} – 2kx + (k – 1) = 0\) has roots \(\alpha \) and \(\beta \) such that \({\alpha ^2} + {\beta ^2} = 4\). Without solving the equation, find the possible values of the real number \(k\).
▶️Answer/Explanation
Markscheme
\(\alpha + \beta = 2k\) A1
\(\alpha \beta = k – 1\) A1
\({(\alpha + \beta )^2} = 4{k^2} \Rightarrow {\alpha ^2} + {\beta ^2} + 2\underbrace {\alpha \beta }_{k – 1} = 4{k^2}\) (M1)
\({\alpha ^2} + {\beta ^2} = 4{k^2} – 2k + 2\)
\({\alpha ^2} + {\beta ^2} = 4 \Rightarrow 4{k^2} – 2k – 2 = 0\) A1
attempt to solve quadratic (M1)
\(k = 1,{\text{ }} – \frac{1}{2}\) A1
[6 marks]