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- A quadratic equation has the form \( ax^2+bx+c=0 \), where \(a,b,c\in\mathbb{R}\) and \(a\ne0\).
- There are always two roots in \( \mathbb{C} \) (distinct real, repeated real, or complex conjugate).
- Methods of solution:
- Factorisation.
- Quadratic formula \( x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} \).
- Completing the square.
- Substitution (e.g. to remove surds) when appropriate.
- Graphical methods (the \(x\)-intercepts of the parabola \(y=ax^2+bx+c\)).
Solving Quadratic Equations by Factorising
Use factors to solve:
(i) \(x^{2} - 5x - 6 = 0\)
(ii) \(y^{2} - 5y = 0\)
(iii) \(4t^{2} - 100 = 0\)
(i) \(x^{2} - 5x - 6 = 0\)
(ii) \(y^{2} - 5y = 0\)
(iii) \(4t^{2} - 100 = 0\)
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Show worked solution
(i)
\(x^{2} - 5x - 6 = 0 \Rightarrow (x - 6)(x + 1) = 0\)
So \(x = 6\) or \(x = -1\).
(ii)
\(y^{2} - 5y = 0 \Rightarrow y(y - 5) = 0\)
So \(y = 0\) or \(y = 5\).
(iii)
\(4t^{2} - 100 = 0 \Rightarrow 4(t^{2} - 25) = 0\)
\(t^{2} - 25 = 0 \Rightarrow (t - 5)(t + 5) = 0\)
So \(t = 5\) or \(t = -5\).
Worked Example: Solving \(3x^{2}+10x-8=0\) by Factorising (Area Model)
▼
Summary
Use the area-model with guide number \(-24x^{2}\) to produce \((3x-2)(x+4)=0\); hence \(x=\tfrac{2}{3}\) or \(x=-4\).
Solving Equations Involving Fractions
Solve (with \(x\neq 0\)):
\[
x - 6 \;=\; \frac{3}{x}
\]
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Show worked solution
Multiply both sides by \(x\): \(x(x-6)=3\).
Rearrange: \(x^{2}-6x-3=0\).
Quadratic formula:
\(x=\dfrac{6\pm\sqrt{36+12}}{2}=\dfrac{6\pm\sqrt{48}}{2}\).
Simplify: \(x=3\pm2\sqrt{3}\) (both non-zero, so valid).
Solving Quadratic Equations with Substitution
Solve for \(x\in\mathbb{R}\):
\[
x^{4}+x^{2}-6=0
\]
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Show worked solution
Let \(y=x^{2}\). Then \(y^{2}+y-6=0\).
Factorise: \((y+3)(y-2)=0\Rightarrow y=-3 \text{ or } y=2\).
Back-substitute \(x^{2}=y\): \(x^{2}=-3\) (no real roots), \(x^{2}=2\Rightarrow x=\pm\sqrt{2}\).
Real solutions: \(x=\sqrt{2},\;x=-\sqrt{2}\).
Solving a Quadratic Surd Equation using Substitution
Solve for \(x\in\mathbb{R}\):
\[
2x + 3\sqrt{x} = 5
\]
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Show worked solution
Let \(y=\sqrt{x}\) (so \(y\ge 0\)), hence \(x=y^{2}\).
Substitute: \(2y^{2}+3y=5 \;\Rightarrow\; 2y^{2}+3y-5=0\).
Quadratic formula:
\[
y=\frac{-3\pm\sqrt{3^{2}-4(2)(-5)}}{2\cdot 2}
=\frac{-3\pm\sqrt{49}}{4}
=\frac{-3\pm 7}{4}.
\]
So \(y=1\) or \(y=-2.5\). Since \(y\ge 0\), discard \(y=-2.5\).
Thus \(\sqrt{x}=1 \Rightarrow x=1\).
Solution: \(x=1\).
- The discriminant of \( ax^2 + bx + c = 0 \) is \( \Delta = b^2 - 4ac \).
- The discriminant determines the nature of the roots:
- \( \Delta > 0 \): two distinct real roots.
- \( \Delta = 0 \): one repeated real root.
- \( \Delta < 0 \): two complex conjugate (non-real) roots.
- If \( \Delta \) is a perfect square, the roots are rational; if \( \Delta > 0 \) but not a perfect square, the roots are irrational (but real).
Using the Discriminant to Find the Nature of Quadratic Roots
Evaluate the discriminant of each of the following, stating whether the equation has
(i) two distinct real roots, (ii) two identical real roots, or (iii) no real roots.
(a) \(3x^{2}+5x-1=0\) (b) \(49x^{2}+42x+9=0\)
(c) \(2x^{2}+8x+9=0\) (d) \(2x^{2}+7x+4=0\)
(a) \(3x^{2}+5x-1=0\) (b) \(49x^{2}+42x+9=0\)
(c) \(2x^{2}+8x+9=0\) (d) \(2x^{2}+7x+4=0\)
Recall the discriminant: \(\Delta=b^{2}-4ac\) for \(ax^{2}+bx+c=0\).
-
(a) \(a=3,\; b=5,\; c=-1\).
\(\Delta=5^{2}-4(3)(-1)=25+12=37>0\).
Two distinct real roots (irrational). -
(b) \(a=49,\; b=42,\; c=9\).
\(\Delta=42^{2}-4(49)(9)=1764-1764=0\).
Two identical real roots (a repeated root). -
(c) \(a=2,\; b=8,\; c=9\).
\(\Delta=8^{2}-4(2)(9)=64-72=-8<0\).
No real roots. -
(d) \(a=2,\; b=7,\; c=4\).
\(\Delta=7^{2}-4(2)(4)=49-32=17>0\).
Two distinct real roots.
Irish specification connection: discriminant determines the number and nature of roots and whether the quadratic graph cuts the \(x\)-axis (two, one, or zero intercepts).
Problem Involving Equal Roots
Find the values of \(k\) so that
\[
-8 + kx - 2x^{2} = 0
\]
has equal real roots.
Rearrange to standard form: \(-2x^{2}+kx-8=0\).
For equal roots, the discriminant \(\Delta=b^{2}-4ac=0\).
Here \(a=-2,\; b=k,\; c=-8\).
\(\Rightarrow\; k^{2}-4(-2)(-8)=0 \;\Longrightarrow\; k^{2}-64=0\).
Hence \(k=\pm 8\).
Values of \(k\): \(k=8\) or \(k=-8\).
Proving Roots are Reals and Rational
Given the equation \(px^{2}+(p+q)x+q=0\) with \(p,q\in\mathbb{R}\) and \(p\neq0\).
(i) Show that the roots are real for all \(p,q\in\mathbb{R}\).
(ii) Show that the roots simplify to rational expressions in \(p\) and \(q\) (hence are rational numbers whenever \(p,q\in\mathbb{Q}\)).
(iii) Hence find (a) the roots in terms of \(p,q\); (b) the factors in terms of \(p,q\).
(i) Show that the roots are real for all \(p,q\in\mathbb{R}\).
(ii) Show that the roots simplify to rational expressions in \(p\) and \(q\) (hence are rational numbers whenever \(p,q\in\mathbb{Q}\)).
(iii) Hence find (a) the roots in terms of \(p,q\); (b) the factors in terms of \(p,q\).
(i) Reality of roots. Here \(a=p,\; b=p+q,\; c=q\). The discriminant is
\[
\Delta=b^{2}-4ac=(p+q)^{2}-4pq=p^{2}-2pq+q^{2}=(p-q)^{2}\ge0.
\]
Hence the equation has real roots for all \(p,q\in\mathbb{R}\) (with \(p\ne0\) so it is quadratic).
(ii) Rational form. Using the quadratic formula,
\[
x=\frac{-(p+q)\pm\sqrt{\Delta}}{2p}=\frac{-(p+q)\pm|p-q|}{2p}.
\]
Splitting by the sign gives
\[
x_1=\frac{-p-q+(p-q)}{2p}=\frac{-2q}{2p}=-\frac{q}{p},\qquad
x_2=\frac{-p-q-(p-q)}{2p}=\frac{-2p}{2p}=-1.
\]
Thus the roots simplify to the rational expressions \(-1\) and \(-\dfrac{q}{p}\); in particular, the roots are rational numbers whenever \(p,q\in\mathbb{Q}\).
(iii)(a) Roots. \(\boxed{x=-1,\;\; x=-\dfrac{q}{p}}\).
(iii)(b) Factors. Hence
\[
px^{2}+(p+q)x+q=p(x+1)\Big(x+\frac{q}{p}\Big)=(x+1)(px+q).
\]
Check: expand \((x+1)(px+q)=px^{2}+(p+q)x+q\) ✓
- A simultaneous system may include one quadratic and one linear equation in two variables.
- The solutions correspond to the intersection points of the parabola (quadratic) and the line (linear).
- Methods of solution:
- Substitution: express one variable from the linear equation and substitute into the quadratic.
- Elimination: eliminate one variable to obtain a single equation in the other when convenient.
- The number of real solutions (0, 1, or 2) depends on how the line meets the parabola.
Quadratic–Linear Simultaneous Equations (Points of Intersection)
Find the point(s) of intersection between the curve \(y=x^{2}+5x+1\) and each line:
(i) \(x-y=-1\) (ii) \(x-y=3\).
(i) \(x-y=-1\) (ii) \(x-y=3\).
(i) Line \(x-y=-1\)
Rewrite the line: \(y=x+1\).
Intersect with the curve: \(x^{2}+5x+1=x+1 \;\Rightarrow\; x^{2}+4x=0\).
Factorise: \(x(x+4)=0 \Rightarrow x=0\) or \(x=-4\).
Find \(y\): If \(x=0\), \(y=0+1=1\); if \(x=-4\), \(y=-4+1=-3\).
Points: \((0,1)\) and \((-4,-3)\).
(ii) Line \(x-y=3\)
Rewrite the line: \(y=x-3\).
Intersect with the curve: \(x^{2}+5x+1=x-3 \;\Rightarrow\; x^{2}+4x+4=0\).
Factorise: \((x+2)^{2}=0 \Rightarrow x=-2\).
Find \(y\): \(y=-2-3=-5\).
Point: \((-2,-5)\).
Quadratic–Linear Simultaneous: No Real Intersection
Show that there are no point(s) of intersection between the line \(x-y=5\) and the curve \(y=x^{2}+5x+1\).
From the line \(x-y=5\) we get \(y=x-5\).
Substitute into the curve: \(x-5=x^{2}+5x+1\).
Rearrange: \(x^{2}+4x+6=0\).
Discriminant \(\Delta=b^{2}-4ac=4^{2}-4(1)(6)=16-24=-8<0\).
Since \(\Delta<0\), there are no real solutions for \(x\); therefore the line and the curve do not meet.
Alternative: \((x+2)^{2}+2=0\) has no real solutions.
- Algebraic methods can be used to solve many real-life problems.
- When an unknown quantity can be represented by a variable and the relationship written as a linear or quadratic equation, the resulting equation can be solved using known algebraic techniques.
- Typical contexts include geometry (perimeter, area, or Pythagoras), motion, and intersection problems (e.g. a line meeting a circle or parabola).
Quadratic & Linear Simultaneous Equations — Context Problem
A right-angled triangle is to be made from a rope 24 m long. If the hypotenuse must be
10 m, find:
(i) an equation (in \(x\) and \(y\)) for the perimeter,
(ii) an equation (in \(x\) and \(y\)) for the hypotenuse,
(iii) solve to find possible lengths of the base \(x\) and height \(y\).
(i) an equation (in \(x\) and \(y\)) for the perimeter,
(ii) an equation (in \(x\) and \(y\)) for the hypotenuse,
(iii) solve to find possible lengths of the base \(x\) and height \(y\).
(i) Perimeter equation: \(x+y+\text{hypotenuse}=24 \Rightarrow x+y+10=24\) ⇒ \(x+y=14\).
(ii) Hypotenuse equation (Pythagoras): \(x^{2}+y^{2}=10^{2}=100\).
(iii) Solve: Square the perimeter: \((x+y)^{2}=196= x^{2}+2xy+y^{2}\). Using \(x^{2}+y^{2}=100\): \(100+2xy=196 \Rightarrow xy=48\).
So \(x\) and \(y\) are the roots of \(t^{2}-14t+48=0\).
Solve: \(t=\dfrac{14\pm\sqrt{14^{2}-4(48)}}{2}= \dfrac{14\pm\sqrt{196-192}}{2}= \dfrac{14\pm2}{2}\Rightarrow t=8\) or \(t=6\).
Possible dimensions: \((x,y)=(8,6)\) or \((6,8)\) (metres).
Check: \(8+6+10=24\) and \(8^{2}+6^{2}=64+36=100\) ✓.
Line & Circle — Will the Paths Cross?
A satellite travels along the line \(x-y=3\).
A comet moves on the curve \(x^{2}+y^{2}-36x+224=0\).
Decide whether their paths will cross.
From the line \(x-y=3\) ⇒ \(y=x-3\).
Substitute into the circle:
\(x^{2}+(x-3)^{2}-36x+224=0\).
Expand:
\(x^{2}+x^{2}-6x+9-36x+224=0 \Rightarrow 2x^{2}-42x+233=0\).
Discriminant:
\(D=b^{2}-4ac=(-42)^{2}-4(2)(233)=1764-1864=-100<0\).
Since \(D<0\), there are no real solutions ⇒
the line and circle do not intersect. The paths will not cross.
Geometric view: Completing the square gives
\((x-18)^{2}+y^{2}=100\) (circle centre \((18,0)\), radius \(10\)); the line \(y=x-3\) stays
outside this circle.
- If \( \alpha \) and \( \beta \) are the roots of a quadratic, the equation can be written as \( (x-\alpha)(x-\beta)=0 \).
- Expanding gives \( x^2-(\alpha+\beta)x+\alpha\beta=0 \).
- Hence:
- Sum of roots \( \alpha+\beta=-\tfrac{b}{a} \)
- Product of roots \( \alpha\beta=\tfrac{c}{a} \)
- If only the sum \(S\) and product \(P\) of the roots are known, the quadratic equation is \( x^2-Sx+P=0 \).
Key Concept — Forming a Quadratic Equation from its Roots
Given roots \(r_1\) and \(r_2\), the quadratic equation can be written as:
\(x^{2} - (r_{1} + r_{2})x + r_{1}r_{2} = 0\)
\(x^{2} - (\text{sum of the roots})x + (\text{product of the roots}) = 0\)
Forming an Equation from Given Roots
Write an equation of a curve whose roots are \(7\) and \(-5\).
Method 1 — Factor form
Roots \(7\) and \(-5\) ⇒ factors \((x-7)\) and \((x+5)\).
Equation: \((x-7)(x+5)=0\).
Expand: \(x^{2}-7x+5x-35=0 \Rightarrow x^{2}-2x-35=0\).
One suitable curve: \(y=x^{2}-2x-35\).
Method 2 — Sum & product of roots
Let \(r_1=7,\; r_2=-5\). Then \(\text{sum}=r_1+r_2=2\), \(\text{product}=r_1r_2=-35\).
Use the identity \(x^{2}-x(r_1+r_2)+r_1r_2=0\).
So \(x^{2}-2x-35=0\) ⇒ \(y=x^{2}-2x-35\).
Finding Coefficients from Given Roots
If \(x = \sqrt{3}\) and \(x = -\frac{\sqrt{3}}{2}\) are the roots of a quadratic equation
\(ax^{2} + bx + c = 0\), find possible values for \(a,\; b,\) and \(c\).
Given roots \(r_1 = \sqrt{3}\) and \(r_2 = -\frac{\sqrt{3}}{2}\).
Sum of roots \(r_1 + r_2 = \sqrt{3} - \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}\).
Product of roots \(r_1r_2 = \sqrt{3} \times \left(-\frac{\sqrt{3}}{2}\right) = -\frac{3}{2}.\)
For a quadratic \(ax^2 + bx + c = 0\),
\(\text{sum} = -\frac{b}{a}\) and \(\text{product} = \frac{c}{a}\).
Thus \(-\frac{b}{a} = \frac{\sqrt{3}}{2}\) and \(\frac{c}{a} = -\frac{3}{2}\).
Taking \(a = 2\) (to remove denominators),
\(b = -\sqrt{3}\) and \(c = -3.\)
Therefore: \(a = 2,\; b = -\sqrt{3},\; c = -3.\)
Equation: \(2x^2 - \sqrt{3}x - 3 = 0.\)
Quadratic from Given Roots
Form a quadratic equation with roots \(x=-3\) and \(x=2\).
- Any quadratic can be written in vertex form \(y=a(x-h)^2+k\) by completing the square.
- The vertex is \((h,k)\); the axis of symmetry is \(x=h\).
- If \(a>0\), the parabola opens upwards and has a minimum value \(k\) at \(x=h\); if \(a<0\), it opens downwards and has a maximum value \(k\) at \(x=h\).
- Completing the square for \(ax^2+bx+c\) gives \[ ax^2+bx+c = a\!\left(x+\frac{b}{2a}\right)^{\!2} + c - \frac{b^{2}}{4a}. \]
- Sign/roots connection (using discriminant \(\Delta=b^2-4ac\)):
- If \(a>0\) and \(\Delta<0\), then \(ax^2+bx+c>0\) for all \(x\).
- If \(a<0\) and \(\Delta<0\), then \(ax^2+bx+c<0\) for all \(x\).
Completing the Square — An Introduction
Solve by completing the square:
\[
x^{2}+10x=39
\]
Step 1: To complete the square, add \(25\) to both sides.
Step 2: Factor the left side as a perfect square: \((x+5)^{2}=64.\)
Step 3: Take the square root: \(x+5=\pm8.\)
Step 4: \(x=3\) or \(x=-13.\)
Check: both values satisfy \(x^{2}+10x=39.\)
Finding the Minimum Value of a Quadratic
Write the quadratic \(x^{2}+4x+1\) in the form \((x-p)^{2}+q\) and hence:
(i) find the minimum point and minimum value of \(x^{2}+4x+1\);
(ii) solve \(x^{2}+4x+1=0\) leaving your answers in surd form.
(i) find the minimum point and minimum value of \(x^{2}+4x+1\);
(ii) solve \(x^{2}+4x+1=0\) leaving your answers in surd form.
Complete the square
\(x^{2}+4x+1=(x^{2}+4x+4)-4+1=(x+2)^{2}-3\).
So in the form \((x-p)^{2}+q\): \(p=-2,\; q=-3\).
(i) Minimum point and value
Vertex of \(y=(x+2)^{2}-3\) is at \(x=-2\).
Minimum value is \(y_{\min}=-3\) at the point \((-2,-3)\).
(ii) Solve \(x^{2}+4x+1=0\) in surd form
\((x+2)^{2}-3=0 \Rightarrow (x+2)^{2}=3\).
\(x+2=\pm\sqrt{3}\Rightarrow x=-2\pm\sqrt{3}\).
Roots: \(x=-2+\sqrt{3}\) and \(x=-2-\sqrt{3}\).
Quadratic from a Graph (Vertex form → Standard form)
(i) Write the equation of the provided graph in the form
\(~y=q-a(x-p)^{2}~\), where \((p,q)\) is the maximum point and \(a\) is a constant.
(ii) By choosing any suitable point on the curve, find \(a\).
(iii) Hence write the equation in the form \(~y=ax^{2}+bx+c\).
(ii) By choosing any suitable point on the curve, find \(a\).
(iii) Hence write the equation in the form \(~y=ax^{2}+bx+c\).
(i) Vertex (maximum) form
From the graph, read the vertex \((p,q)\) (the highest point).
Write \(y=q-a(x-p)^{2}\). The minus sign ensures a downward opening parabola.
(ii) Find \(a\) from any point on the curve
Pick a convenient point \((x_1,y_1)\) on the graph (not the vertex).
Substitute into \(y=q-a(x-p)^{2}\): \(y_1=q-a(x_1-p)^{2}\).
Solve for \(a\): \(\displaystyle a=\frac{q-y_1}{(x_1-p)^{2}}\).
Use clear integer/reading points from the grid to avoid rounding error.
(iii) Expand to standard form \(y=ax^{2}+bx+c\)
Start with \(y=q-a(x-p)^{2}=q-a(x^{2}-2px+p^{2})\).
Hence \(y=(-a)x^{2}+(2ap)x+(q-ap^{2})\).
Therefore \(a_{\text{std}}=-a,\quad b=2ap,\quad c=q-ap^{2}\).
Insert your numeric \(p,q,a\) from parts (i)-(ii) to get \(y=ax^{2}+bx+c\).
- A surd is an irrational root (e.g. \( \sqrt{2},\sqrt{5} \)) that cannot be expressed exactly as a fraction.
- Surds are left in exact form unless a decimal approximation is required.
- Rules:
- \( \sqrt{a}\times\sqrt{b}=\sqrt{ab},\qquad \dfrac{\sqrt{a}}{\sqrt{b}}=\sqrt{\dfrac{a}{b}} \)
- Simplify \( \sqrt{a^2b}=a\sqrt{b} \) only when \( a>0 \).
- Rationalise denominators by multiplying by a suitable form of 1 to remove the surd from the denominator.
Simplifying Surds
(i) Express \(\sqrt{80}\) in the form \(a\sqrt{5}\), where \(a\) is an integer.
(ii) Express \((4 - \sqrt{5})^{2}\) in the form \(b + c\sqrt{5}\), where \(b\) and \(c\) are integers.
(ii) Express \((4 - \sqrt{5})^{2}\) in the form \(b + c\sqrt{5}\), where \(b\) and \(c\) are integers.
(i)
\(\sqrt{80} = \sqrt{16 \times 5} = \sqrt{16}\sqrt{5} = 4\sqrt{5}\).
Answer: \(a = 4 \Rightarrow \sqrt{80} = 4\sqrt{5}\).
(ii)
\((4 - \sqrt{5})^{2} = 4^{2} - 2(4)(\sqrt{5}) + (\sqrt{5})^{2}\)
\(= 16 - 8\sqrt{5} + 5 = 21 - 8\sqrt{5}\).
Answer: \(b = 21,~ c = -8 \Rightarrow (4 - \sqrt{5})^{2} = 21 - 8\sqrt{5}\).
Simplifying Surd Fractions (Dividing by Surds)
Simplify:
(i) \(\dfrac{\sqrt{12}}{5\sqrt{3} - \sqrt{27}}\)
(ii) \(\dfrac{7}{\sqrt{13} - \sqrt{11}}\)
(i) \(\dfrac{\sqrt{12}}{5\sqrt{3} - \sqrt{27}}\)
(ii) \(\dfrac{7}{\sqrt{13} - \sqrt{11}}\)
(i)
\(\sqrt{12} = 2\sqrt{3}\) and \(\sqrt{27} = 3\sqrt{3}\).
\(\dfrac{\sqrt{12}}{5\sqrt{3} - \sqrt{27}} = \dfrac{2\sqrt{3}}{5\sqrt{3} - 3\sqrt{3}} = \dfrac{2\sqrt{3}}{2\sqrt{3}} = 1.\)
Answer: \(1\).
(ii)
Multiply top and bottom by the conjugate \((\sqrt{13} + \sqrt{11})\):
\(\dfrac{7}{\sqrt{13} - \sqrt{11}} \times \dfrac{\sqrt{13} + \sqrt{11}}{\sqrt{13} + \sqrt{11}} = \dfrac{7(\sqrt{13} + \sqrt{11})}{13 - 11}\).
Simplify denominator: \(13 - 11 = 2\).
\(\Rightarrow \dfrac{7}{2}(\sqrt{13} + \sqrt{11}) = \dfrac{7\sqrt{13}}{2} + \dfrac{7\sqrt{11}}{2}.\)
Answer: \(\dfrac{7\sqrt{13}}{2} + \dfrac{7\sqrt{11}}{2}\).
- To solve equations containing surds, first isolate a square root, then square both sides.
- Domain matters: ensure each radicand is non-negative and respect any sign conditions introduced (e.g. a side declared non-negative).
- Squaring can introduce extraneous solutions; always check in the original equation.
- Useful identities:
- \( \sqrt{a}\sqrt{b}=\sqrt{ab}, \quad \dfrac{\sqrt{a}}{\sqrt{b}}=\sqrt{\dfrac{a}{b}} \) for \(a,b\ge0,\ b\ne0\)
- \( \sqrt{a^2b}=a\sqrt{b} \) for \(a\ge0\)
- Rationalise denominators by multiplying by a conjugate, e.g. \( \dfrac{1}{p+q\sqrt{r}} \cdot \dfrac{p-q\sqrt{r}}{p-q\sqrt{r}} \).
Key Concept — Solving Surd Equations
For one surd:
- Isolate the surd term.
- Square both sides to remove the square root.
- Solve the resulting equation and check each solution.
For two surds:
- Isolate one surd on one side first.
- Square both sides to remove that surd.
- Simplify — you’ll still have one surd left.
- Isolate and square again to remove the second surd.
- Solve and verify every solution by substitution.
Rule of thumb: isolate → square → simplify → repeat → check.
Solving a Surd Equation and Verifying Solutions
Solve \(x=\sqrt{x+6}\), and verify any solution(s).
Start with \(x=\sqrt{x+6}\).
Square both sides: \(x^{2}=x+6\).
Rearrange: \(x^{2}-x-6=0\).
Factorise: \((x-3)(x+2)=0\).
Hence \(x=3\) or \(x=-2\).
Check both:
For \(x=3:\) LHS \(=3,\) RHS \(=\sqrt{3+6}=\sqrt{9}=3\) ✓ valid.
For \(x=-2:\) LHS \(=-2,\) RHS \(=\sqrt{-2+6}=\sqrt{4}=2\) ✗ not valid.
For \(x=3:\) LHS \(=3,\) RHS \(=\sqrt{3+6}=\sqrt{9}=3\) ✓ valid.
For \(x=-2:\) LHS \(=-2,\) RHS \(=\sqrt{-2+6}=\sqrt{4}=2\) ✗ not valid.
Final solution: \(x=3\).
Solving an Equation with Two Surd Terms (and Verifying)
Solve \(\sqrt{5x+6}-\sqrt{2x}=2\), and verify all solutions.
Domain: \(5x+6\ge0\) and \(2x\ge0 \Rightarrow x\ge0\).
Move one surd: \(\sqrt{5x+6}=2+\sqrt{2x}\).
Square: \(5x+6=4+4\sqrt{2x}+2x\Rightarrow 3x+2=4\sqrt{2x}\).
Square again: \((3x+2)^{2}=16\cdot2x\Rightarrow 9x^{2}-20x+4=0\).
Solve: discriminant \(=400-144=256\Rightarrow x=\dfrac{20\pm16}{18}\Rightarrow x=2\) or \(x=\dfrac{2}{9}\).
Verification in the original equation:
\(x=2:\ \sqrt{16}-\sqrt{4}=4-2=2\ \checkmark\)
\(x=\dfrac{2}{9}:\ \sqrt{\dfrac{64}{9}}-\sqrt{\dfrac{4}{9}}=\dfrac{8}{3}-\dfrac{2}{3}=2\ \checkmark\)
\(x=2:\ \sqrt{16}-\sqrt{4}=4-2=2\ \checkmark\)
\(x=\dfrac{2}{9}:\ \sqrt{\dfrac{64}{9}}-\sqrt{\dfrac{4}{9}}=\dfrac{8}{3}-\dfrac{2}{3}=2\ \checkmark\)
Solution set: \(\{\,2,\ \dfrac{2}{9}\,\}\).
Squaring can introduce extraneous roots; always verify in the original equation.
- Remainder Theorem: For a polynomial \(f(x)\), the remainder on division by \(x − a\) is \(f(a)\).
- Factor Theorem: \(x − a\) is a factor of \(f(x)\) iff \(f(a)=0\).
- Once a linear factor is found, divide to reduce the degree, then factorise the quotient (by inspection, completing the square, or using the quadratic formula).
- Strategy for integer roots: test \(\pm\) factors of the constant term (or \(\pm\) factors of \(\tfrac{\text{constant}}{\text{leading coefficient}}\) for non-monic polynomials).
Key Concept — The Factor Theorem
If f(k) = 0, then (x − k) is a factor of f(x).
Conversely, if (x − k) is a factor, then f(k) = 0.
More generally, if (a x − k) is a factor, then f( k ⁄ a ) = 0.
Using the Factor Theorem
Show that \((2x-3)\) is a factor of \(2x^{3}-5x^{2}+5x-3\).
Let \(f(x)=2x^{3}-5x^{2}+5x-3\). If \((2x-3)\) is a factor, then \(x=\tfrac{3}{2}\) should be a root.
Evaluate \(f\!\left(\tfrac{3}{2}\right)\):
\[ 2\!\left(\tfrac{3}{2}\right)^{3}-5\!\left(\tfrac{3}{2}\right)^{2} +5\!\left(\tfrac{3}{2}\right)-3 =\tfrac{27}{4}-\tfrac{45}{4}+\tfrac{15}{2}-3 =\tfrac{27-45+30-12}{4}=0. \] Hence \(f\!\left(\tfrac{3}{2}\right)=0\) and by the Factor Theorem \((2x-3)\) is a factor.
\[ 2\!\left(\tfrac{3}{2}\right)^{3}-5\!\left(\tfrac{3}{2}\right)^{2} +5\!\left(\tfrac{3}{2}\right)-3 =\tfrac{27}{4}-\tfrac{45}{4}+\tfrac{15}{2}-3 =\tfrac{27-45+30-12}{4}=0. \] Hence \(f\!\left(\tfrac{3}{2}\right)=0\) and by the Factor Theorem \((2x-3)\) is a factor.
Find the other factor. Write
\[
2x^{3}-5x^{2}+5x-3=(2x-3)(ax^{2}+bx+c).
\]
Matching coefficients gives \(a=1,\; b=-1,\; c=1\). Therefore
\[
2x^{3}-5x^{2}+5x-3=(2x-3)(x^{2}-x+1).
\]
Check: expand \((2x-3)(x^{2}-x+1)=2x^{3}-5x^{2}+5x-3\) ✓.
Note that \(x^{2}-x+1\) has discriminant \(1-4=-3<0\), so the only real root is \(x=\tfrac{3}{2}\).
Using the Factor Theorem to Find Unknown Coefficients
If \((x-2)\) and \((x+1)\) are both factors of \(ax^{3}+3x^{2}-9x+b\), find the values of \(a\) and \(b\).
Let \(f(x)=ax^{3}+3x^{2}-9x+b\).
Since \((x-2)\) is a factor, \(f(2)=0\):
\[
8a+12-18+b=0 \;\Rightarrow\; 8a-6+b=0 \;\Rightarrow\; b=6-8a.
\]
Since \((x+1)\) is a factor, \(f(-1)=0\):
\[
-a+3+9+b=0 \;\Rightarrow\; -a+12+b=0 \;\Rightarrow\; b=a-12.
\]
Equate the two expressions for \(b\):
\[
6-8a=a-12 \;\Rightarrow\; 18=9a \;\Rightarrow\; \boxed{a=2}.
\]
Then \(b=a-12=2-12=\boxed{-10}\).
Check / factorisation. With \(a=2,\;b=-10\),
\[
2x^{3}+3x^{2}-9x-10=(x-2)(x+1)(2x+5).
\]
Expanding verifies the coefficients, so the values are correct.
Factor Theorem: \(x-r\) is a factor of \(f(x)\) iff \(f(r)=0\).
Factorising Cubic Expressions
Factorise the cubic expression \(f(x)=2x^{3}+x^{2}-13x+6\).
Search for a rational root. Try \(x=2\):
\[
f(2)=2(8)+4-26+6=16+4-26+6=0.
\]
Hence \(x=2\) is a root and \((x-2)\) is a factor.
Divide by \((x-2)\) to find the quadratic factor. Synthetic division on
\(2,\;1,\;-13,\;6\) gives the quotient \(2x^{2}+5x-3\).
Factor the quadratic:
\[
2x^{2}+5x-3=(2x-1)(x+3).
\]
Complete factorisation:
\[
\boxed{\,2x^{3}+x^{2}-13x+6=(x-2)(2x-1)(x+3)\, }.
\]
Check: expand \((x-2)(2x-1)(x+3)\) to recover \(2x^{3}+x^{2}-13x+6\) ✓.
Solving a Cubic Equation
Solve the equation \(2x^{3}-4x^{2}-22x+24=0\).
Factor out \(2\): \(2\big(x^{3}-2x^{2}-11x+12\big)=0\).
Rational root test. Try \(x=1\): \(1-2-11+12=0\Rightarrow x=1\) is a root \(\Rightarrow (x-1)\) is a factor.
Divide \(x^{3}-2x^{2}-11x+12\) by \((x-1)\) to get the quotient \(x^{2}-x-12\).
Factor the quadratic: \(x^{2}-x-12=(x-4)(x+3)\).
Complete factorisation:
\[
2x^{3}-4x^{2}-22x+24=2(x-1)(x-4)(x+3).
\]
Solutions: \(x=1,\; x=4,\; x=-3\).
- A polynomial of degree \( n \) has the form \( f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 \), where \( a_n\ne0 \).
- Zeros (roots) \( r_1,r_2,\dots \) satisfy \( f(r_i)=0 \); the factorised form is \( f(x)=a(x-r_1)(x-r_2)\cdots(x-r_n) \).
- To find a polynomial from its roots and a known point, substitute the point to determine the leading coefficient \( a \).
- Graphs of higher-degree polynomials are continuous and smooth; the number of real roots corresponds to the number of \(x\)-intercepts.
Key Concept — Finding a Polynomial from its Graph
The roots of a polynomial tell you where the graph cuts or touches the x-axis.
The function can be written as a multiple of its factors:
\(f(x) = k(x - r_1)(x - r_2)(x - r_3)\dots\)
The constant \(k\) (the scale factor) can be found by substituting a known point on the graph, e.g. \((x_0, y_0)\).
Finding an Expression for a Cubic Polynomial from its Graph
By examining the graph, find an expression for this cubic polynomial.
From the graph, the cubic cuts the \(x\)-axis at \(x=-2,\;1,\;3\).
Hence the function can be written as
\[
f(x)=k(x+2)(x-1)(x-3),
\]
where \(k\) is a constant.
Substitute a known point on the graph, e.g. \((0,6)\):
\[
6=k(0+2)(0-1)(0-3)=k(2)(-1)(-3)=6k.
\]
Therefore \(k=1\).
Final expression:
\[
\boxed{f(x)=(x+2)(x-1)(x-3)}.
\]
This method links algebraic roots with the graph’s intercepts and vertical scale.
Finding the Expression for a Degree-4 Polynomial from its Graph
The graph of the polynomial \(f(x)=a(x+b)(x+c)(x+d)(x+d)\) is shown.
Find the values of \(a,b,c,d\).
Find the values of \(a,b,c,d\).
Read roots from the graph.
Let the \(x\)-intercepts be \(x=r_1,\; r_2,\; r_3\) with a touch/turn at \(x=r_3\) (double root).
Then
\[
f(x)=k\,(x-r_1)(x-r_2)(x-r_3)^{2}.
\]
Comparing with \(a(x+b)(x+c)(x+d)^{2}\) gives
\[
b=-r_1,\qquad c=-r_2,\qquad d=-r_3,\qquad a=k.
\]
Use the \(y\)-intercept to find \(a=k\).
Read off \(f(0)=y_0\) from the graph (the point where the curve meets the \(y\)-axis). Then
\[
y_0=f(0)=k(-r_1)(-r_2)(-r_3)^{2}
\;\;\Rightarrow\;\;
k=\frac{y_0}{(-r_1)(-r_2)(-r_3)^{2}}.
\]
Hence the values.
\[
a=\frac{y_0}{(-r_1)(-r_2)(-r_3)^{2}},\qquad
b=-r_1,\quad c=-r_2,\quad d=-r_3.
\]
Substitute the numerical values \(r_1, r_2, r_3, y_0\) as read from your diagram.
Notes: a “bounce”/tangent at an intercept indicates an even multiplicity (here 2).
The leading coefficient \(a\) controls vertical stretch and end behaviour.
Extra Practice Questions
Solution.
The general cubic with these roots is \( f(x) = a(x-1)(x-2)(x-3) \).
Substitute \( (4, 24) \): \( 24 = a(3)(2)(1) \Rightarrow a = 4 \).
Hence \( f(x) = 4(x-1)(x-2)(x-3) \).
Solution.
The general quadratic is \( f(x) = a(x+1)(x-3) \).
Substitute \( (0, -6) \): \( -6 = a(1)(-3) \Rightarrow a = 2 \).
Therefore \( f(x) = 2(x+1)(x-3) \).
