The laws of indices for all rational exponents, including a⁰, negative powers and a^(m/n) as a root

Simplifying surds and multiplying out brackets containing surds

Rationalising denominators of the form √b and a±√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 b²-4ac and the nature of the roots, including finding an unknown constant for equal roots

Disguised quadratics in √x, x^(1/3), 2ˣ or x²

1. This page is the algebra every later pure topic leans on: powers written as xⁿ, exact surd answers, and quadratics that you solve, sketch or test with the discriminant.

2. aᵐ×aⁿ=aᵐ⁺ⁿ, aᵐ÷aⁿ=aᵐ⁻ⁿ, (aᵐ)ⁿ=aᵐⁿ, a⁰=1, a⁻ⁿ=1/aⁿ, a^(m/n)=(ⁿ√a)ᵐ.

3. √(ab)=√a√b: pull out the largest square factor, so √72=6√2. But √(a+b) is not √a+√b.

4. To rationalise 1/(a+√b), multiply top and bottom by a-√b; the denominator becomes a²-b.

5. 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.

6. b²-4ac>0: two distinct real roots; =0: one repeated root (the curve touches the x-axis); <0: no real roots.

7. If an equation contains a power and its square (√x and x, 2ˣ and 4ˣ), substitute u and solve a quadratic in u.

The three basic laws come from counting factors: a³×a² 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: a³÷a³=a⁰ and also =1, so a⁰=1 (for a≠0).

Negative: a⁰÷aⁿ=a⁻ⁿ, so a⁻ⁿ=1/aⁿ. A negative power means "reciprocal", not "negative number": 2⁻³=⅛.

Unit fractions: (a^(1/n))ⁿ=a¹, so a^(1/n) is the nth root of a: 8^(1/3)=2, x^(1/2)=√x.

General fractions: a^(m/n)=(ⁿ√a)ᵐ=ⁿ√(aᵐ). Take the root first and the numbers stay small: 27^(2/3)=3²=9.

The laws only combine powers of the same base. To compare 4ˣ with 8ʸ, first write both as powers of 2.

These are not in the formulae booklet: learn them. Also (ab)ⁿ=aⁿbⁿ and (a/b)⁻ⁿ=(b/a)ⁿ. Take a>0 whenever the power is a fraction.

WORKED EXAMPLE · QUESTION

(a) Find the exact value of (16/81)^(-3/4). (b) Simplify fully ((9x⁴)^(3/2))/(3x)², where x>0.

SOLUTION

(a) The negative power turns the fraction upside down: (16/81)^(-3/4)=(81/16)^(3/4). Fourth root first, then cube: (81/16)^(1/4)=(3/2), and (3/2)³= (27/8)

(b) Apply the power to each factor of the numerator: (9x⁴)^(3/2)=9^(3/2)x^(4×3/2)=27x⁶, and (3x)²=9x². So 27x⁶/9x²= 3x⁴

A common slip in (b) is 9^(3/2)=9×(3/2). The power applies to the 9 as well: √9=3, then 3³=27.

WORKED EXAMPLE · QUESTION

(a) Write (5x²-3√x)/2x√x in the form Axᵖ+Bx^q, where A, B, p and q are constants. (b) Solve 4ˣ×2ˣ⁺¹=32.

SOLUTION

(a) Write every root as a power: x√x=x^(3/2) and √x=x^(1/2). Then divide each term separately, subtracting the powers: 5x²/(2x^(3/2))-(3x^(1/2))/(2x^(3/2))= (5/2) x^(1/2)-(3/2) x⁻¹

(b) Write everything as a power of 2: 2²ˣ×2ˣ⁺¹=2⁵, so 2³ˣ⁺¹=2⁵. Equal bases, so equal powers: 3x+1=5 and x=(4/3)

This "same base" method only works when every term can be written as a power of one number. Equations such as 3ˣ=7 need logarithms (P2).

A surd is an irrational root such as √2, √12 or ∛5. Leaving an answer as a surd keeps it exact; a decimal is only an approximation.

The key rules are √(ab)=√a×√b and √(a/b)=√a/√b, together with √a×√a=a. To simplify, split off the largest square factor: √75=√(25×3)=5√3. "Simplest form" means the number left under the root has no square factor other than 1.

Like surds collect like algebra: 5√3+4√3=9√3. Unlike surds such as √2+√3 cannot be combined. Brackets expand exactly like algebra, using √a×√a=a to tidy up.

WORKED EXAMPLE · QUESTION

(a) Simplify √75+√48-√12, giving your answer in the form k√3. (b) Expand and simplify (4-√3)(2+3√3).

SOLUTION

(a) Largest square factor of each number: √75=5√3, √48=4√3, √12=2√3. So 5√3+4√3-2√3= 7√3

(b) Four products, as with any two brackets: 8+12√3-2√3-3×3=8+10√3-9= -1+10√3

The last product is (-√3)(3√3)=-3×(√3)²=-9. Forgetting that √3×√3 is a whole number is the usual source of error.

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 √b: multiply top and bottom by √b, because √b×√b=b. So 6/√3=6√3/3=2√3.

Denominator a+√b: multiply top and bottom by the conjugate a-√b. The difference of two squares removes the surd: (a+√b)(a-√b)=a²-b.

The same works for √a±√b: (√a+√b)(√a-√b)=a-b.

WORKED EXAMPLE · QUESTION

Write (7+√5)/(3-√5) in the form p+q√5, where p and q are rational numbers.

SOLUTION

Step 1 Multiply top and bottom by the conjugate 3+√5: ((7+√5)(3+√5))/((3-√5)(3+√5)).

Step 2 Denominator by difference of two squares: 3²-(√5)²=9-5=4. Numerator: 21+7√5+3√5+5=26+10√5.

Step 3 Divide each term by 4: (26+10√5)/4= (13/2)+(5/2)√5

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).

The graph of y=ax²+bx+c (a≠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 ax²+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.

Completing the square rewrites x²+bx+c as (x+b/2)²-(b/2)²+c: halve the coefficient of x, then subtract the square you have added. The picture shows why: x²+6x is a square with two strips of 3x, and it needs an extra 3²=9 to make the full square (x+3)².

If a≠1, take a out as a factor of the x² 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.

So the vertex is at x=-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 ax²+bx+c=0 is how the quadratic formula is derived.

WORKED EXAMPLE · QUESTION

f(x)=2x²-12x+7. (a) Write f(x) in the form a(x+p)²+q. (b) State the coordinates of the vertex of y=f(x) and the equation of its line of symmetry. (c) Solve f(x)=0, giving your answers in simplest surd form.

SOLUTION

(a) Take out the factor 2 from the first two terms: 2(x²-6x)+7. Complete the square inside the bracket: 2[(x-3)²-9]+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=(11/2) and x-3=±√(11/2). Since √(11/2)=√11/√2=√22/2, x=3±√22/2

As decimals the roots are 0.655 and 5.35: symmetrical about x=3, as they must be.

The quadratic formula is not in the formulae booklet: learn it by heart. The expression under the root, b²-4ac, is the discriminant (next section).

WORKED EXAMPLE · QUESTION

(a) Solve 6x²+x-15=0. (b) By completing the square, solve x²-6x+4=0, giving exact answers. (c) Solve 3x²+4x-2=0, giving your answers in the form (p±√q)/3 and also to 3 significant figures.

SOLUTION

(a) ac=-90; the factor pair 10 and −9 sums to 1. 6x²+10x-9x-15=2x(3x+5)-3(3x+5)=(2x-3)(3x+5), so x=(3/2) or x=-(5/3)

(b) (x-3)²-9+4=0, so (x-3)²=5 and x=3±√5

(c) Formula with a=3, b=4, c=-2: x=(-4±√(16+24))/6=(-4±2√10)/6= (-2±√10)/3; x=0.387 or x=-1.72 (3 s.f.)

In (c), -4ac=-4×3×(-2)=+24. Simplify √40=2√10, then cancel the common factor 2 from every term.

In the formula, ±√(b²-4ac) gives two different values only if b²-4ac is positive. This quantity is the discriminant.

b²-4ac>0: two distinct real roots — the graph crosses the x-axis twice.

b²-4ac=0: one repeated (equal) root, x=-b/2a — the vertex sits on the x-axis.

b²-4ac<0: no real roots — the graph never meets the x-axis.

"Real roots" (with no "distinct") means b²-4ac≥0. The discriminant is not in the formulae booklet: learn it. Write the equation as ax²+bx+c=0 before reading off a, b and c, including their signs.

WORKED EXAMPLE · QUESTION

The equation x²+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.

SOLUTION

(a) a=1, b=k, c=k+3. Equal roots: k²-4(k+3)=0, so k²-4k-12=0, (k-6)(k+2)=0 and k=6 or k=-2

(b) k=6: x²+6x+9=(x+3)², so x=-3. k=-2: x²-2x+1=(x-1)², so x=1

(c) The discriminant k²-4k-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.

A disguised quadratic has the shape a(something)²+b(something)+c=0. Common "somethings" are √x (because x=(√x)²), x^(1/3), x², and 2ˣ (because 4ˣ=(2ˣ)²).

Method: let u = the something, rewrite, solve for u, then convert back to x. Reject impossible values: √x and 2ˣ are never negative, and x²=-4 has no real solution. Say why you reject a value. Watch powers like 2²ˣ⁺¹=2×2²ˣ=2u²: split the index with the laws of indices first.

WORKED EXAMPLE · QUESTION

Solve (a) 2x-7√x+3=0, (b) 2²ˣ⁺¹-17×2ˣ+8=0, (c) x⁴-5x²-36=0, for real values of x.

SOLUTION

(a) Let u=√x, so x=u²: 2u²-7u+3=0, (2u-1)(u-3)=0, u=½ or 3. Then √x=½ or √x=3, so x=¼ or x=9

(b) 2²ˣ⁺¹=2×(2ˣ)², so with u=2ˣ: 2u²-17u+8=0, (2u-1)(u-8)=0, u=½ or 8. Then 2ˣ=2⁻¹ or 2ˣ=2³, so x=-1 or x=3

(c) Let u=x²: u²-5u-36=0, (u-9)(u+4)=0, u=9 or u=-4. x²=9 gives x=±3; x²=-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 √x≥0. Substituting x=¼: 2×¼-7×½+3=0.

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 2²ˣ-5×2ˣ+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+½ at² 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.

1. Find the exact value of 32^(-3/5).

2. Simplify √200-√18.

3. Rationalise the denominator of 1/(√7-2).

4. Find the coordinates of the vertex of y=x²+10x+19.

5. The equation px²-4x+1=0, where p≠0, has equal roots. Find p.

6. Solve x^(2/3)-5x^(1/3)+4=0.

1. 32^(1/5)=2, so 32^(3/5)=2³=8 and 32^(-3/5)=⅛.

2. √200=√(100×2)=10√2 and √18=√(9×2)=3√2, so the answer is 7√2.

3. Multiply top and bottom by √7+2: the denominator becomes 7-4=3, so the answer is (√7+2)/3.

4. x²+10x+19=(x+5)²-25+19=(x+5)²-6, so the vertex (a minimum) is (-5, -6).

5. b²-4ac=16-4p=0, so p=4. Then 4x²-4x+1=(2x-1)², with repeated root x=½.

6. Let u=x^(1/3): u²-5u+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.)