Use substitution to form and solve a quadratic equation in order to solve a related equation

Solving Quadratics

• Factorisation (when possible).
• Completing the square.
• Quadratic formula \[ x=\frac{-b\pm\sqrt{\Delta}}{2a}. \]

Worked Example – Vertex by Differentiation

Find the vertex and axis of symmetry of \(y=3x^{2}-12x+7\).

1. Derivative: \(\displaystyle \frac{dy}{dx}=6x-12\).
2. Set derivative to zero: \(6x-12=0\Rightarrow x=2\) (axis of symmetry).
3. Substitute back: \(y=3(2)^{2}-12(2)+7=12-24+7=-5\).
4. Vertex: \((2,-5)\).

Quadratic Inequalities – Systematic Approach

Two equivalent methods are required by the syllabus:

1. Sign‑chart method (factorise, locate critical points, test intervals).
2. Graphical method (draw \(y=ax^{2}+bx+c\); the solution set is where the curve lies above or below the \(x\)-axis, depending on the inequality).

Example – Solving a Quadratic Inequality

Solve \(x^{2}-4x-5\le0\).

1. Factor: \((x-5)(x+1)\le0\).
2. Critical points: \(-1\) and \(5\).
3. Sign chart:

   Interval	Test point	Sign of \((x-5)(x+1)\)
   \((-\infty,-1)\)	\(x=-2\)	\(+\)
   \((-1,5)\)	\(x=0\)	\(-\)
   \((5,\infty)\)	\(x=6\)	\(+\)

4. Since we need “\(\le0\)”, the solution is the interval where the sign is negative or zero: \(\boxed{-1\le x\le5}\).

3. Factors of Polynomials

• Remainder Theorem: The remainder when \(f(x)\) is divided by \((x-a)\) is \(f(a)\).
• Factor Theorem: \((x-a)\) is a factor of \(f(x)\) iff \(f(a)=0\).
• Use synthetic division or long division once a root is known.

Worked Example – Cubic Factorisation

Factorise \(f(x)=x^{3}-6x^{2}+11x-6\).

1. Test possible integer roots \(\pm1,\pm2,\pm3,\pm6\). \(f(1)=0\) ⇒ \((x-1)\) is a factor.
2. Synthetic division by \((x-1)\) gives \(x^{2}-5x+6\).
3. Factor the quadratic: \((x-2)(x-3)\).
4. Result: \(f(x)=(x-1)(x-2)(x-3)\).

Practice

Factor \(x^{4}-5x^{2}+4\) completely.

4. Equations, Inequalities and Graphs

4.1 Absolute‑Value Equations & Inequalities

• \(|ax+b|=c\;(c\ge0)\) → \(ax+b=c\) or \(ax+b=-c\).
• \(|ax+b|=cx+d\) (linear right‑hand side) → split into two cases, remembering that the right‑hand side must be non‑negative in the case chosen.
• For inequalities, translate \(|ax+b|c\) into \(ax+b-c\).

Example

Solve \(|2x-5|=7\).

\(2x-5=7\Rightarrow x=6\) or \(2x-5=-7\Rightarrow x=-1\). Both satisfy the original equation.

4.2 Quadratic‑Type Equations & Inequalities

These are equations that become quadratic after a suitable substitution, typically \(t=x^{n}\) with \(n=2\) or \(n=3\).

4.3 Using Substitution to Form and Solve a Quadratic

General Procedure

1. Identify a repeated or cumbersome part of the expression and set it equal to a new variable \(t\).
2. Rewrite the whole equation in terms of \(t\).
3. Arrange into standard quadratic form \(at^{2}+bt+c=0\).
4. Solve for \(t\) (factorisation, completing the square, or quadratic formula).
5. Back‑substitute the original expression for \(t\) and solve the resulting simpler equations.
6. Check every solution in the original equation to discard extraneous roots.

Worked Example

Solve \(\displaystyle \frac{1}{x}+\frac{1}{x^{2}}=6\) for real \(xeq0\).

1. Substitution: Let \(t=\dfrac{1}{x}\); then \(t^{2}=\dfrac{1}{x^{2}}\).
2. Rewrite: \(t+t^{2}=6\;\Longrightarrow\;t^{2}+t-6=0\).
3. Solve the quadratic: \[ t^{2}+t-6=(t+3)(t-2)=0\;\Longrightarrow\;t=-3\;\text{or}\;t=2. \]
4. Back‑substitute:
   • \(t=-3\Rightarrow \dfrac{1}{x}=-3\Rightarrow x=-\dfrac13\).
   • \(t=2\Rightarrow \dfrac{1}{x}=2\Rightarrow x=\dfrac12\).
5. Check (both satisfy the original equation).

Summary Table

Additional Practice (Substitution)

1. \(x^{4}-5x^{2}+4=0\) (set \(t=x^{2}\)).
2. \(\sqrt{2x+3}+\sqrt{x-1}=5\) (set \(t=\sqrt{x-1}\)).
3. \(\displaystyle \frac{x}{x-2}+\frac{x-2}{x}=3\) (set \(t=x+\frac{2}{x}\) or clear denominators first).

4.4 Solving Quadratic‑Type Inequalities

Two complementary strategies are required:

1. Sign‑chart method (factorise, locate zeros, test intervals).
2. Graphical method (draw the parabola; the solution set is the region where the curve lies above or below the \(x\)-axis, according to the inequality sign).

Example

Solve \(2x^{2}-7x+3>0\).

1. Factor: \((2x-1)(x-3)>0\).
2. Critical points: \(x=\tfrac12,\;3\).
3. Sign chart shows the product is positive for \((-\infty,\tfrac12)\) and \((3,\infty)\).
4. Solution: \(\displaystyle x<\frac12\ \text{or}\ x>3\).

4.5 Sketching Cubic Polynomials

1. Find real roots (Factor Theorem or calculator).
2. Determine multiplicities – even multiplicity → tangent; odd multiplicity → cross.
3. Compute turning points by solving \(f'(x)=0\) (a quadratic).
4. Plot intercepts and turning points; sketch a smooth “S‑shaped” curve.

Worked Example

Sketch \(y=x^{3}-3x^{2}-4x+12\).

1. Roots: test \(x=1\Rightarrow0\) ⇒ factor \((x-1)\). Division gives \(x^{2}-2x-12=(x-4)(x+3)\). Roots: \(-3,1,4\).
2. Derivative: \(y'=3x^{2}-6x-4\). Solve \(3x^{2}-6x-4=0\) → \(x=1\pm\frac{\sqrt{13}}{3}\) (turning points).
3. Plot \((-3,0),\;(1,0),\;(4,0)\) and the two turning points; join smoothly.

5. Simultaneous Equations

• Linear – linear: substitution or elimination.
• Linear – quadratic (or other non‑linear): isolate one variable, substitute into the other, then solve the resulting quadratic (or higher‑degree) equation.

Worked Example

Solve \(\begin{cases} y-x+3=0 \\ x^{2}-3xy+y^{2}+19=0 \end{cases}\).

1. From the first equation, \(y=x-3\).
2. Substitute into the second: \[ x^{2}-3x(x-3)+(x-3)^{2}+19=0\Longrightarrow -x^{2}+3x+28=0. \]
3. Multiply by \(-1\): \(x^{2}-3x-28=0\) → \((x-7)(x+4)=0\). Hence \(x=7\) or \(x=-4\).
4. Back‑substitute: \(y=7-3=4\) or \(y=-4-3=-7\).
5. Solutions: \((7,4)\) and \((-4,-7)\).

Practice

Solve \(\begin{cases} 2x+y=5 \\ x^{2}+y^{2}=13 \end{cases}\).

6. Logarithmic & Exponential Functions

• Exponential form: \(y=a^{x}\) with \(a>0,\;aeq1\).
• Logarithmic form: \(y=\log_{a}x\) is the inverse of \(a^{y}=x\).
• Key laws:
  • \(a^{m}a^{n}=a^{m+n}\)
  • \(\dfrac{a^{m}}{a^{n}}=a^{m-n}\)
  • \((a^{m})^{n}=a^{mn}\)
  • \(\log_{a}(mn)=\log_{a}m+\log_{a}n\)
  • \(\log_{a}\dfrac{m}{n}=\log_{a}m-\log_{a}n\)
  • \(\log_{a}(m^{n})=n\log_{a}m\)
• Change‑of‑base formula: \(\displaystyle \log_{a}b=\frac{\log_{c}b}{\log_{c}a}\) (usually \(c=10\) or \(e\)).

Worked Example – Solving an Exponential Equation

Solve \(3^{2x-1}=27\).

1. Express the right‑hand side with the same base: \(27=3^{3}\).
2. Equate exponents: \(2x-1=3\).
3. Solve: \(2x=4\Rightarrow x=2\).

Worked Example – Solving a Logarithmic Equation

Solve \(\log_{2}(x+3)=4\).

1. Rewrite in exponential form: \(2^{4}=x+3\).
2. Compute: \(16=x+3\Rightarrow x=13\).

Practice

1. Solve \(5^{x}=125\).
2. Solve \(\log_{10}(2x-1)=2\).
3. Use change‑of‑base to evaluate \(\log_{3}20\) (give answer to 3 dp).