P1.1 Indices, surds and quadratics
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This is a P1 (AS) sub-topic. Simultaneous equations and quadratic inequalities are on the P1 equations and inequalities page, transformations of graphs are on the P1 curve sketching page, and exponential equations that need logarithms belong to P2.
What this page covers
8 points
- The laws of indices for all rational exponents, including \(a^0\), negative powers and \(a^{m/n}\) as a root
- Simplifying surds and multiplying out brackets containing surds
- Rationalising denominators of the form \(\sqrt b\) and \(a\pm\sqrt b\)
- Quadratic functions and their graphs: shape, intercepts, line of symmetry and vertex
- Completing the square, and reading off the vertex and the minimum or maximum value
- Solving quadratic equations by factorising, by completing the square and by the formula
- The discriminant \(b^2-4ac\) and the nature of the roots, including finding an unknown constant for equal roots
- Disguised quadratics in \(\sqrt x\), \(x^{1/3}\), \(2^x\) or \(x^2\)
In one minute
7 points
- This page is the algebra every later pure topic leans on: powers written as \(x^n\), exact surd answers, and quadratics that you solve, sketch or test with the discriminant.
- \(a^m\times a^n=a^{m+n}\), \(a^m\div a^n=a^{m-n}\), \((a^m)^n=a^{mn}\), \(a^0=1\), \(a^{-n}=1/a^n\), \(a^{m/n}=(\sqrt[n]{a})^m\).
- \(\sqrt{ab}=\sqrt a\sqrt b\): pull out the largest square factor, so \(\sqrt{72}=6\sqrt2\). But \(\sqrt{a+b}\) is not \(\sqrt a+\sqrt b\).
- To rationalise \(\dfrac{1}{a+\sqrt b}\), multiply top and bottom by \(a-\sqrt b\); the denominator becomes \(a^2-b\).
- \(a(x+p)^2+q\) has vertex \((-p, q)\) and line of symmetry \(x=-p\); a minimum if \(a>0\), a maximum if \(a<0\).
- \(b^2-4ac>0\): two distinct real roots; \(=0\): one repeated root (the curve touches the \(x\)-axis); \(<0\): no real roots.
- If an equation contains a power and its square (\(\sqrt x\) and \(x\), \(2^x\) and \(4^x\)), substitute \(u\) and solve a quadratic in \(u\).
Contents
14 sections
- How do the laws of indices work for negative and fractional powers?
- What is a surd, and how do you simplify one?
- How do you rationalise a denominator?
- What does the graph of a quadratic tell you?
- How does completing the square find the vertex?
- Which method should you use to solve a quadratic equation?
- How does the discriminant decide the number of real roots?
- How do you spot and solve a disguised quadratic?
- Exam tips for P1.1
- Common misconceptions
- How this topic connects
- Key formulae
- Check your understanding
- Specification checklist
How do the laws of indices work for negative and fractional powers?
The three basic laws come from counting factors: \(a^3\times a^2\) has \(3+2\) factors of \(a\). Insisting that the same laws hold for every rational power forces the meaning of zero, negative and fractional powers.
- Zero: \(a^3\div a^3=a^0\) and also \(=1\), so \(a^0=1\) (for \(a\ne 0\)).
- Negative: \(a^0\div a^n=a^{-n}\), so \(a^{-n}=1/a^n\). A negative power means "reciprocal", not "negative number": \(2^{-3}=\tfrac18\).
- Unit fractions: \((a^{1/n})^n=a^1\), so \(a^{1/n}\) is the \(n\)th root of \(a\): \(8^{1/3}=2\), \(x^{1/2}=\sqrt x\).
- General fractions: \(a^{m/n}=(\sqrt[n]{a})^m=\sqrt[n]{a^m}\). Take the root first and the numbers stay small: \(27^{2/3}=3^2=9\).
The laws only combine powers of the same base. To compare \(4^x\) with \(8^y\), first write both as powers of 2.
\[a^m a^n=a^{m+n}\qquad a^m\div a^n=a^{m-n}\qquad (a^m)^n=a^{mn}\] \[a^0=1\qquad a^{-n}=\frac{1}{a^n}\qquad a^{m/n}=(\sqrt[n]{a})^m\]These are not in the formulae booklet: learn them. Also \((ab)^n=a^nb^n\) and \(\left(\tfrac{a}{b}\right)^{-n}=\left(\tfrac{b}{a}\right)^n\). Take \(a>0\) whenever the power is a fraction.
(a) Find the exact value of \(\left(\dfrac{16}{81}\right)^{-3/4}\). (b) Simplify fully \(\dfrac{(9x^4)^{3/2}}{(3x)^2}\), where \(x>0\).
(a) The negative power turns the fraction upside down: \(\left(\tfrac{16}{81}\right)^{-3/4}=\left(\tfrac{81}{16}\right)^{3/4}\). Fourth root first, then cube: \(\left(\tfrac{81}{16}\right)^{1/4}=\tfrac32\), and \(\left(\tfrac32\right)^3=\) \(\tfrac{27}{8}\)
(b) Apply the power to each factor of the numerator: \((9x^4)^{3/2}=9^{3/2}x^{4\times 3/2}=27x^6\), and \((3x)^2=9x^2\). So \(\dfrac{27x^6}{9x^2}=\) \(3x^4\)
A common slip in (b) is \(9^{3/2}=9\times\tfrac32\). The power applies to the 9 as well: \(\sqrt9=3\), then \(3^3=27\).
(a) Write \(\dfrac{5x^2-3\sqrt x}{2x\sqrt x}\) in the form \(Ax^p+Bx^q\), where \(A\), \(B\), \(p\) and \(q\) are constants. (b) Solve \(4^x\times 2^{x+1}=32\).
(a) Write every root as a power: \(x\sqrt x=x^{3/2}\) and \(\sqrt x=x^{1/2}\). Then divide each term separately, subtracting the powers: \(\dfrac{5x^2}{2x^{3/2}}-\dfrac{3x^{1/2}}{2x^{3/2}}=\) \(\tfrac52 x^{1/2}-\tfrac32 x^{-1}\)
(b) Write everything as a power of 2: \(2^{2x}\times 2^{x+1}=2^5\), so \(2^{3x+1}=2^5\). Equal bases, so equal powers: \(3x+1=5\) and \(x=\tfrac43\)
This "same base" method only works when every term can be written as a power of one number. Equations such as \(3^x=7\) need logarithms (P2).
What is a surd, and how do you simplify one?
A surd is an irrational root such as \(\sqrt2\), \(\sqrt{12}\) or \(\sqrt[3]{5}\). Leaving an answer as a surd keeps it exact; a decimal is only an approximation.
The key rules are \(\sqrt{ab}=\sqrt a\times\sqrt b\) and \(\sqrt{\tfrac{a}{b}}=\dfrac{\sqrt a}{\sqrt b}\), together with \(\sqrt a\times\sqrt a=a\). To simplify, split off the largest square factor: \(\sqrt{75}=\sqrt{25\times 3}=5\sqrt3\). "Simplest form" means the number left under the root has no square factor other than 1.
Like surds collect like algebra: \(5\sqrt3+4\sqrt3=9\sqrt3\). Unlike surds such as \(\sqrt2+\sqrt3\) cannot be combined. Brackets expand exactly like algebra, using \(\sqrt a\times\sqrt a=a\) to tidy up.
(a) Simplify \(\sqrt{75}+\sqrt{48}-\sqrt{12}\), giving your answer in the form \(k\sqrt3\). (b) Expand and simplify \((4-\sqrt3)(2+3\sqrt3)\).
(a) Largest square factor of each number: \(\sqrt{75}=5\sqrt3\), \(\sqrt{48}=4\sqrt3\), \(\sqrt{12}=2\sqrt3\). So \(5\sqrt3+4\sqrt3-2\sqrt3=\) \(7\sqrt3\)
(b) Four products, as with any two brackets: \(8+12\sqrt3-2\sqrt3-3\times 3=8+10\sqrt3-9=\) \(-1+10\sqrt3\)
The last product is \((-\sqrt3)(3\sqrt3)=-3\times(\sqrt3)^2=-9\). Forgetting that \(\sqrt3\times\sqrt3\) is a whole number is the usual source of error.
How do you rationalise a denominator?
Rationalising means rewriting a fraction so that its denominator has no surd. The value does not change: you multiply by a form of 1.
- Denominator \(\sqrt b\): multiply top and bottom by \(\sqrt b\), because \(\sqrt b\times\sqrt b=b\). So \(\dfrac{6}{\sqrt3}=\dfrac{6\sqrt3}{3}=2\sqrt3\).
- Denominator \(a+\sqrt b\): multiply top and bottom by the conjugate \(a-\sqrt b\). The difference of two squares removes the surd: \((a+\sqrt b)(a-\sqrt b)=a^2-b\).
- The same works for \(\sqrt a\pm\sqrt b\): \((\sqrt a+\sqrt b)(\sqrt a-\sqrt b)=a-b\).
Write \(\dfrac{7+\sqrt5}{3-\sqrt5}\) in the form \(p+q\sqrt5\), where \(p\) and \(q\) are rational numbers.
Step 1 Multiply top and bottom by the conjugate \(3+\sqrt5\): \(\dfrac{(7+\sqrt5)(3+\sqrt5)}{(3-\sqrt5)(3+\sqrt5)}\).
Step 2 Denominator by difference of two squares: \(3^2-(\sqrt5)^2=9-5=4\). Numerator: \(21+7\sqrt5+3\sqrt5+5=26+10\sqrt5\).
Step 3 Divide each term by 4: \(\dfrac{26+10\sqrt5}{4}=\) \(\tfrac{13}{2}+\tfrac52\sqrt5\)
Check with a calculator: both the original and the answer are about 12.090. A numerical check catches sign slips in the conjugate.
Using your calculator
- Type the original expression and your simplified answer separately; they must give the same decimal.
- Many scientific calculators display surds exactly: use this to check, but write every line of working yourself.
- For powers like \(27^{2/3}\), enter the index in brackets: 27^(2÷3).
Multiplying only the denominator by the conjugate. Top and bottom must both be multiplied, or the value of the fraction changes. A second trap: \((3-\sqrt5)(3+\sqrt5)\) is \(9-5=4\), not \(9+5\).
What does the graph of a quadratic tell you?
The graph of \(y=ax^2+bx+c\) (\(a\ne 0\)) is a parabola: ∪-shaped if \(a>0\), ∩-shaped if \(a<0\).
It crosses the \(y\)-axis at \((0, c)\), and it meets the \(x\)-axis where \(ax^2+bx+c=0\), so the roots of the equation are the \(x\)-intercepts of the graph. It is symmetrical about a vertical line through its vertex (turning point). When there are two roots, the line of symmetry is halfway between them.
A sketch shows the correct shape, with every point where the curve meets the axes labelled, and the vertex when it is asked for. It does not need to be to scale.
For a ∩-shaped curve, factorise with the negative sign handled first: \(8+2x-x^2=-(x^2-2x-8)=-(x-4)(x+2)\), which is the same as \((4-x)(2+x)\). Check the \(y\)-intercept from your factors: \(4\times 2=8\).
How does completing the square find the vertex?
Completing the square rewrites \(x^2+bx+c\) as \(\left(x+\tfrac{b}{2}\right)^2-\left(\tfrac{b}{2}\right)^2+c\): halve the coefficient of \(x\), then subtract the square you have added. The picture shows why: \(x^2+6x\) is a square with two strips of \(3x\), and it needs an extra \(3^2=9\) to make the full square \((x+3)^2\).
If \(a\ne 1\), take \(a\) out as a factor of the \(x^2\) and \(x\) terms first, complete the square inside the bracket, then multiply back out. In the form \(a(x+p)^2+q\), the squared term is never negative, so for \(a>0\) the least value is \(q\), when \(x=-p\). The vertex is \((-p, q)\) and the line of symmetry is \(x=-p\). For \(a<0\) the same point is a maximum, with greatest value \(q\).
\[ax^2+bx+c=a\left(x+\frac{b}{2a}\right)^2+c-\frac{b^2}{4a}\]So the vertex is at \(x=-\dfrac{b}{2a}\), which is also the line of symmetry. This is not in the formulae booklet: learn the method, not the formula. Rearranging this form for \(ax^2+bx+c=0\) is how the quadratic formula is derived.
\(\mathrm{f}(x)=2x^2-12x+7\). (a) Write \(\mathrm{f}(x)\) in the form \(a(x+p)^2+q\). (b) State the coordinates of the vertex of \(y=\mathrm{f}(x)\) and the equation of its line of symmetry. (c) Solve \(\mathrm{f}(x)=0\), giving your answers in simplest surd form.
(a) Take out the factor 2 from the first two terms: \(2(x^2-6x)+7\). Complete the square inside the bracket: \(2\left[(x-3)^2-9\right]+7=2(x-3)^2-18+7=\) \(2(x-3)^2-11\)
(b) Minimum point \((3, -11)\); line of symmetry \(x=3\)
(c) \(2(x-3)^2=11\), so \((x-3)^2=\tfrac{11}{2}\) and \(x-3=\pm\sqrt{\tfrac{11}{2}}\). Since \(\sqrt{\tfrac{11}{2}}=\dfrac{\sqrt{11}}{\sqrt2}=\dfrac{\sqrt{22}}{2}\), \(x=3\pm\dfrac{\sqrt{22}}{2}\)
As decimals the roots are 0.655 and 5.35: symmetrical about \(x=3\), as they must be.
Which method should you use to solve a quadratic equation?
| Method | Use it when | Watch out for |
|---|---|---|
| Factorising | the roots are rational — try it first when the numbers are small | rearrange to "= 0" first; never divide both sides by \(x\) (you lose the root \(x=0\)) |
| Completing the square | the question asks for it, or you also need the vertex; \(a=1\) and \(b\) even makes it quick | both square roots, ± |
| Quadratic formula | it will not factorise, or a decimal answer is wanted | bracket a negative \(b\) when squaring; divide the whole numerator by \(2a\) |
The quadratic formula is not in the formulae booklet: learn it by heart. The expression under the root, \(b^2-4ac\), is the discriminant (next section).
(a) Solve \(6x^2+x-15=0\). (b) By completing the square, solve \(x^2-6x+4=0\), giving exact answers. (c) Solve \(3x^2+4x-2=0\), giving your answers in the form \(\dfrac{p\pm\sqrt q}{3}\) and also to 3 significant figures.
(a) \(ac=-90\); the factor pair 10 and −9 sums to 1. \(6x^2+10x-9x-15=2x(3x+5)-3(3x+5)=(2x-3)(3x+5)\), so \(x=\tfrac32\) or \(x=-\tfrac53\)
(b) \((x-3)^2-9+4=0\), so \((x-3)^2=5\) and \(x=3\pm\sqrt5\)
(c) Formula with \(a=3\), \(b=4\), \(c=-2\): \(x=\dfrac{-4\pm\sqrt{16+24}}{6}=\dfrac{-4\pm 2\sqrt{10}}{6}=\) \(\dfrac{-2\pm\sqrt{10}}{3}\); \(x=0.387\) or \(x=-1.72\) (3 s.f.)
In (c), \(-4ac=-4\times 3\times(-2)=+24\). Simplify \(\sqrt{40}=2\sqrt{10}\), then cancel the common factor 2 from every term.
How does the discriminant decide the number of real roots?
In the formula, \(\pm\sqrt{b^2-4ac}\) gives two different values only if \(b^2-4ac\) is positive. This quantity is the discriminant.
- \(b^2-4ac>0\): two distinct real roots — the graph crosses the \(x\)-axis twice.
- \(b^2-4ac=0\): one repeated (equal) root, \(x=-\dfrac{b}{2a}\) — the vertex sits on the \(x\)-axis.
- \(b^2-4ac<0\): no real roots — the graph never meets the \(x\)-axis.
"Real roots" (with no "distinct") means \(b^2-4ac\ge 0\). The discriminant is not in the formulae booklet: learn it. Write the equation as \(ax^2+bx+c=0\) before reading off \(a\), \(b\) and \(c\), including their signs.
The equation \(x^2+kx+(k+3)=0\), where \(k\) is a constant, has equal roots. (a) Find the possible values of \(k\). (b) For each value, find the repeated root. (c) State the set of values of \(k\) for which the equation has no real roots.
(a) \(a=1\), \(b=k\), \(c=k+3\). Equal roots: \(k^2-4(k+3)=0\), so \(k^2-4k-12=0\), \((k-6)(k+2)=0\) and \(k=6\) or \(k=-2\)
(b) \(k=6\): \(x^2+6x+9=(x+3)^2\), so \(x=-3\). \(k=-2\): \(x^2-2x+1=(x-1)^2\), so \(x=1\)
(c) The discriminant \(k^2-4k-12\) is a ∪-shaped quadratic in \(k\) with roots −2 and 6, so it is negative between them: \(-2<k<6\)
Part (c) uses a quadratic inequality; the full method is on the P1 equations and inequalities page.
How do you spot and solve a disguised quadratic?
A disguised quadratic has the shape \(a(\text{something})^2+b(\text{something})+c=0\). Common "somethings" are \(\sqrt x\) (because \(x=(\sqrt x)^2\)), \(x^{1/3}\), \(x^2\), and \(2^x\) (because \(4^x=(2^x)^2\)).
Method: let \(u\) = the something, rewrite, solve for \(u\), then convert back to \(x\). Reject impossible values: \(\sqrt x\) and \(2^x\) are never negative, and \(x^2=-4\) has no real solution. Say why you reject a value. Watch powers like \(2^{2x+1}=2\times 2^{2x}=2u^2\): split the index with the laws of indices first.
Solve (a) \(2x-7\sqrt x+3=0\), (b) \(2^{2x+1}-17\times 2^x+8=0\), (c) \(x^4-5x^2-36=0\), for real values of \(x\).
(a) Let \(u=\sqrt x\), so \(x=u^2\): \(2u^2-7u+3=0\), \((2u-1)(u-3)=0\), \(u=\tfrac12\) or 3. Then \(\sqrt x=\tfrac12\) or \(\sqrt x=3\), so \(x=\tfrac14\) or \(x=9\)
(b) \(2^{2x+1}=2\times(2^x)^2\), so with \(u=2^x\): \(2u^2-17u+8=0\), \((2u-1)(u-8)=0\), \(u=\tfrac12\) or 8. Then \(2^x=2^{-1}\) or \(2^x=2^3\), so \(x=-1\) or \(x=3\)
(c) Let \(u=x^2\): \(u^2-5u-36=0\), \((u-9)(u+4)=0\), \(u=9\) or \(u=-4\). \(x^2=9\) gives \(x=\pm 3\); \(x^2=-4\) has no real solutions, so it is rejected.
In (a) both values of \(u\) are positive, so both are valid. Had a negative \(u\) appeared, it would be rejected because \(\sqrt x\ge 0\). Substituting \(x=\tfrac14\): \(2\times\tfrac14-7\times\tfrac12+3=0\).
Exam tips for P1.1
When a question says "exact", "in simplest surd form" or "in the form \(p+q\sqrt3\)", a decimal answer does not count, however accurate. Keep √ and fractions to the end, and simplify the surd fully.
Rewrite roots and reciprocals as powers, one term at a time: \(\dfrac{3}{\sqrt x}=3x^{-1/2}\) and \(\dfrac{1}{2x^3}=\tfrac12 x^{-3}\). The coefficient stays outside the power.
Write the equation in the form \(ax^2+bx+c=0\), state \(a\), \(b\) and \(c\) (with signs and brackets), write \(b^2-4ac=0\) (or \(>0\), \(<0\)) explicitly, then expand line by line to the given result. The final line is given, so the working is the whole answer.
Factor out \(a\) from the \(x^2\) and \(x\) terms only, complete the square inside, then multiply the subtracted square by \(a\) as you remove the outer bracket. Expand your answer mentally to check it.
Solving for \(u\) is only half the question. Convert every value back to \(x\), reject any that are impossible with a reason, and give all the solutions that remain.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| A negative power gives a negative number: \(4^{-1/2}=-2\). | The minus sign in the index means "reciprocal". | \(4^{-1/2}=1/\sqrt4=\tfrac12\). |
| \(\sqrt{a+b}=\sqrt a+\sqrt b\). | Roots split over multiplication, not addition: \(\sqrt{9+16}=5\), but \(\sqrt9+\sqrt{16}=7\). | \(\sqrt{ab}=\sqrt a\sqrt b\) only. |
| \(3x^{-2}=\dfrac{1}{3x^2}\). | The power belongs to \(x\) only, not to the 3. | \(3x^{-2}=\dfrac{3}{x^2}\); but \((3x)^{-2}=\dfrac{1}{9x^2}\). |
| \((x+4)^2-7\) has its vertex at \((4, -7)\). | The bracket is zero when \(x=-4\). | \(a(x+p)^2+q\) has vertex \((-p, q)\): here \((-4, -7)\). |
| \(b^2-4ac=0\) means there are no roots. | The ± part of the formula vanishes, leaving one value. | Zero discriminant: one repeated root, where the graph touches the axis. |
| Dividing \(x^2=5x\) by \(x\) gives the only answer \(x=5\). | Dividing by \(x\) assumes \(x\ne 0\) and loses a root. | \(x^2-5x=0\), \(x(x-5)=0\), \(x=0\) or 5. |
How this topic connects
P1 Differentiation and integration
Every term must be a power of \(x\) before you differentiate or integrate, so index form (worked example 2) comes first. Stationary points of a quadratic agree with the vertex from completing the square.
P1 Equations, inequalities and curve sketching
The discriminant decides whether a line meets a curve, and the roots found here become the critical values of quadratic inequalities and the intercepts on sketches and transformed graphs.
P2, P3, M1 and IAL Physics
Equations like \(2^{2x}-5\times 2^x+6=0\) return in P2 with answers that need logarithms; in P3 the completed square gives the range of a quadratic function and the domain needed for an inverse. In kinematics, \(s=ut+\tfrac12 at^2\) is a quadratic in \(t\): two positive roots mean a projectile passes the same height twice, and a negative discriminant means it never reaches that height.
Key formulae
| Formula | Use | In the formulae list? |
|---|---|---|
| \(a^ma^n=a^{m+n},\ a^m\div a^n=a^{m-n},\ (a^m)^n=a^{mn}\) | simplifying powers of the same base | No — learn it |
| \(a^0=1,\ a^{-n}=1/a^n,\ a^{m/n}=(\sqrt[n]{a})^m\) | zero, negative and fractional powers | No — learn it |
| \(\sqrt{ab}=\sqrt a\sqrt b,\ \sqrt{a/b}=\sqrt a/\sqrt b\) | simplifying surds | No — learn it |
| \((a+\sqrt b)(a-\sqrt b)=a^2-b\) | rationalising with the conjugate | No — derive it (difference of two squares) |
| \(ax^2+bx+c=a(x+b/2a)^2+c-b^2/4a\) | vertex, line of symmetry, maximum or minimum value | No — derive it by completing the square |
| \(x=\left(-b\pm\sqrt{b^2-4ac}\right)/2a\) | roots of any quadratic equation | No — learn it |
| \(b^2-4ac\) (\(>0,\ =0,\ <0\)) | number of real roots; finding an unknown constant | No — learn it |
Check your understanding
1. Find the exact value of \(32^{-3/5}\).
\(32^{1/5}=2\), so \(32^{3/5}=2^3=8\) and \(32^{-3/5}=\tfrac18\).
2. Simplify \(\sqrt{200}-\sqrt{18}\).
\(\sqrt{200}=\sqrt{100\times 2}=10\sqrt2\) and \(\sqrt{18}=\sqrt{9\times 2}=3\sqrt2\), so the answer is \(7\sqrt2\).
3. Rationalise the denominator of \(\dfrac{1}{\sqrt7-2}\).
Multiply top and bottom by \(\sqrt7+2\): the denominator becomes \(7-4=3\), so the answer is \(\dfrac{\sqrt7+2}{3}\).
4. Find the coordinates of the vertex of \(y=x^2+10x+19\).
\(x^2+10x+19=(x+5)^2-25+19=(x+5)^2-6\), so the vertex (a minimum) is \((-5, -6)\).
5. The equation \(px^2-4x+1=0\), where \(p\ne 0\), has equal roots. Find \(p\).
\(b^2-4ac=16-4p=0\), so \(p=4\). Then \(4x^2-4x+1=(2x-1)^2\), with repeated root \(x=\tfrac12\).
6. Solve \(x^{2/3}-5x^{1/3}+4=0\).
Let \(u=x^{1/3}\): \(u^2-5u+4=0\), \((u-1)(u-4)=0\), so \(u=1\) or 4. Cubing, \(x=1\) or \(x=64\). (Cube roots can be negative, so no value of \(u\) needs to be rejected here.)
Specification checklist
Tick every box before the exam. Wording is ours; numbers refer to the specification sections.
- (P1 1.1) I can use the laws of indices for any rational power, including zero and negative powers, and write roots as fractional powers.
- (P1 1.1) I can rewrite expressions such as \((5x^2-3\sqrt x)/(2x\sqrt x)\) as a sum of powers of \(x\), and solve equations by writing both sides as powers of one base.
- (P1 1.2) I can simplify surds, collect like surds and multiply out brackets that contain surds.
- (P1 1.2) I can rationalise denominators of the form \(\sqrt b\), \(a\pm\sqrt b\) and \(\sqrt a\pm\sqrt b\).
- (P1 1.3) I can sketch the graph of a quadratic function, showing its shape and where it meets the axes, and link the roots of the equation to the \(x\)-intercepts.
- (P1 1.4) I can use the discriminant to decide whether a quadratic has two distinct, equal or no real roots, and find an unknown constant from that condition.
- (P1 1.5) I can complete the square, including when the coefficient of \(x^2\) is not 1, and read off the vertex, line of symmetry and greatest or least value.
- (P1 1.5) I can solve quadratic equations by factorising, completing the square and the formula, and solve disguised quadratics by substitution.