Algebra, Year 11: completing the square and the quadratic formula

5; 25; −18

HintHalve the coefficient of x, then take away the square of that half.

The whyHalf of −10 is −5, giving (x − 5)². That bracket expands to x² − 10x + 25, which is 25 too many, so subtract 25. Then −25 + 7 = −18, so x² − 10x + 7 = (x − 5)² − 18.

Completing the square

HintThe name describes turning x² + 6x into a perfect squared bracket by adding what is missing.

The whyThe form (x + p)² + q shows the turning point of the graph, (−p, q), and lets you solve the equation by taking square roots. It works for every quadratic, including those that do not factorise.

x = −2 + √7 or x = −2 − √7

HintGet the squared bracket on its own before taking square roots.

The whyx² + 4x − 3 = (x + 2)² − 4 − 3 = (x + 2)² − 7. So (x + 2)² = 7, x + 2 = ±√7 and x = −2 ± √7. Both signs of the root are needed, giving two solutions.

(x + 3)² expands to x² + 6x + 9, which has an extra 9 that x² + 6x does not have

HintMultiply the bracket out and compare with what you started from.

The why(x + 3)(x + 3) = x² + 3x + 3x + 9. The x² and 6x are what we want; the 9 is an unwanted extra, so it is taken off again. In general x² + bx = (x + b/2)² − (b/2)².

2(x + 3)² − 13

HintTake the 2 out of the first two terms before completing the square inside the bracket.

The why2x² + 12x + 5 = 2(x² + 6x) + 5 = 2[(x + 3)² − 9] + 5 = 2(x + 3)² − 18 + 5 = 2(x + 3)² − 13. The −9 is inside the bracket, so it is doubled when the bracket is removed.

The quadratic formula

HintIt is named after the type of equation it solves.

The whya, b and c are the numbers in front of x², in front of x, and on their own, read with their signs. AQA expects this formula to be known; it is not given in the exam. It solves every quadratic, including those that do not factorise.

x = −2 + √3 or x = −2 − √3

HintHere a = 1, b = 4 and c = 1; simplify the root at the end.

The whyb² − 4ac = 16 − 4 = 12, so x = (−4 ± √12)/2. Since √12 = 2√3, this is (−4 ± 2√3)/2 = −2 ± √3. Every term on top is divided by the 2.

49

HintTake care with the signs of b and c.

The whya = 2, b = −5, c = −3. b² = (−5)² = 25 and 4ac = 4 × 2 × (−3) = −24, so b² − 4ac = 25 − (−24) = 49. Then x = (5 ± 7)/4, giving x = 3 or x = −1/2.

The quadratic will not factorise neatly, so use the quadratic formula

HintWhole-number or simple fraction answers would not need rounding.

The whyFactorising only finds solutions that are whole numbers or simple fractions. A request for decimal places, significant figures or surd form means the solutions are not that tidy, so the formula (or completing the square) is needed. Here x = (−3 ± √29)/2, which is 1.19 or −4.19.

x = 2.35 or x = −0.85

HintWrite down a, b and c with their signs before substituting.

The whya = 2, b = −3, c = −4, so b² − 4ac = 9 + 32 = 41. Then x = (3 ± √41)/4. Taking + gives 2.3507…, and taking − gives −0.8507….

x = 2, y = 4 or x = −1, y = 1

HintBoth right-hand sides equal y, so set them equal to each other.

The whyx² = x + 2 gives x² − x − 2 = 0, so (x − 2)(x + 1) = 0 and x = 2 or x = −1. Substituting into y = x + 2 gives y = 4 or y = 1. The answers come in pairs.

substitute

HintYou replace a letter with the expression it equals.

The whyFor x + y = 5 and x² + y² = 13, write y = 5 − x and put it in place of y: x² + (5 − x)² = 13. This leaves a quadratic in x alone, which can then be solved. Elimination by adding or subtracting does not work, because the equations are of different types.

3

HintWhere they cross, the two expressions for y are equal.

The whyx² = 2x + 3 gives x² − 2x − 3 = 0, so (x − 3)(x + 1) = 0 and x = 3 or x = −1. The other point is (3, 9). Solving the simultaneous equations is the same as finding where the graphs intersect.

12; 3; 4

HintExpand the bracket, collect terms, then divide everything by 2.

The whyx² + x² + 2x + 1 = 25 gives 2x² + 2x − 24 = 0, and dividing by 2 gives x² + x − 12 = 0. This factorises to (x + 4)(x − 3) = 0. With x = 3, y = 3 + 1 = 4; the other solution is x = −4, y = −3.

The line touches the curve at just one point, (1, 1): it is a tangent

HintCount how many times the two graphs meet.

The whyx² = 2x − 1 gives x² − 2x + 1 = 0, which is (x − 1)² = 0, a repeated root. Two solutions would mean the line cuts the curve twice; one repeated solution means it only touches.