KS4 Mathematics A6 Quadratic equations

A6.1 Solving quadratic equations by factorization Contents A6 Quadratic equations A A6.1 Solving quadratic equations by factorization A A6.2 Completing the square A A6.3 Using the quadratic formula A A6.4 Equations involving algebraic fractions A A6.5 Problems leading to quadratic equations

Find the width of the rectangle The length of a rectangle is 4 cm more than its width. The area of the rectangle is 45 cm2. Find the width of the rectangle. If we call the width of the rectangle x we can draw the following diagram: x x + 4 Using the information about the area of the rectangle we can write an equation: x(x + 4) = 45

Find the width of the rectangle The solution to x(x + 4) = 45 will give us the width of the rectangle. In this example, it should be quite easy to spot that x = 5 is a possible solution to this equation, because 5 × 9 = 45 The width of the rectangle is therefore 5 cm. However, there is another value of x that will also solve the equation x(x + 4) = 45. This is because x(x + 4) = 45 is an example of a quadratic equation.

Quadratic equations This is easier to spot if we multiply out the bracket, x(x + 4) = 45 x2 + 4x = 45 We usually arrange quadratic equations so that all the terms are on the left-hand side of the equals sign, leaving a 0 on the right-hand side. In this example we would have x2 + 4x – 45 = 0 The general form of a quadratic equation is ax2 + bx + c = 0 Where a, b and c are constants and a ≠ 0.

Quadratic equations We can solve the quadratic equation x2 + 4x – 45 = 0 in full by factorizing the expression on the left-hand side. This means that we can write the equation in the form (x + ….)(x + ….) = 0 We need to find two integers that add together to make 4 and multiply together to make –45. Because –45 is negative, one of the numbers must be positive and one must be negative. By considering the factors of 45 we find that the two numbers must be 9 and –5. We can therefore write x2 + 4x – 45 = 0 as (x + 9)(x – 5) = 0

Quadratic equations When two numbers multiply together to make 0, one of the numbers must be 0, so if (x + 9)(x – 5) = 0 we can conclude that either x + 9 = 0 or x – 5 = 0 This gives us two solutions that solve the quadratic equation: x = – 9 and x = 5 In the context of finding the width of a rectangle we cannot allow a negative length, and so x = 5 is the only valid solution. Many other problems that lead to quadratic equations however, would require both solutions.

Solving quadratic equations by factorization Solve the equation x2 = 3x by factorization. Start by rearranging the equation so that the terms are on the left-hand side, x2 – 3x = 0 Factorizing the left-hand side gives us x(x – 3) = 0 x = 0 So or x – 3 = 0 Pupils might be tempted to start by dividing through by x. In a quadratic of this form (where c = 0, there is always the solution x = 0), this leads to the non-zero root, but makes it easy to overlook x = 0. If pupils suggest dividing through, encourage them to factorize instead. x = 3

Solving quadratic equations by factorization Solve the equation x2 – 5x = –4 by factorization. Start by rearranging the equation so that the terms are on the left-hand side. x2 – 5x + 4 = 0 We need to find two integers that add together to make –5 and multiply together to make 4. Because 4 is positive and –5 is negative, both the integers must be negative. These are –1 and –4. Factorizing the left-hand side gives us (x – 1)(x – 4) = 0 x – 1 = 0 or x – 4 = 0 x = 1 x = 4

Solving quadratic equations by factorization It may be a good idea to practice adding and multiplying negative numbers before attempting this activity. See the activity in N1.2 Calculating with integers.

Demonstrating solutions using graphs Use this activity to demonstrate graphically the solutions to quadratic equations that factorize and do not factorize. Reveal the coordinates of the points that intersect the x-axis and relate these to the solution of the equation. Establish that without the use of a computer or a graphics calculator this is not an accurate method of solving the equation. For a more accurate solution we need to use an algebraic method. It is useful, however, for demonstrating the nature of the solutions, for example where there is only one solution, or no solutions. The use of graphs to solve equations in covered in more detail in A9.3

A6.5 Problems leading to quadratic equations Contents A6 Quadratic equations A A6.1 Solving quadratic equations by factorization A A6.2 Completing the square (Not Required) A A6.3 Using the quadratic formula A A6.4 Equations with fractions (Not Required) A A6.5 Problems leading to quadratic equations

A6.3 Using the quadratic formula Contents A6 Quadratic equations A A6.1 Solving quadratic equations by factorization A A6.2 Completing the square A A6.3 Using the quadratic formula A A6.4 Equations involving algebraic fractions A A6.5 Problems leading to quadratic equations

Using the quadratic formula Any quadratic equation of the form, ax2 + bx + c = 0 can be solved by substituting the values of a, b and c into the formula, x = –b ± b2 – 4ac 2a Tell pupils that they are not required to learn this formula since it is given in the examination. It is a complex formula, however, and so stress that some practice is needed to use it correctly. Ask pupils if they can see how this formula will lead to two solutions. More able pupils could be asked to complete the square for the general quadratic equation to derive the formula. This equation can be derived by completing the square on the general form of the quadratic equation.

Using the quadratic formula Use the quadratic formula to solve x2 – 7x + 8 = 0. 1x2 – 7x + 8 = 0 x = –b ± b2 – 4ac 2a x = 2 × 1 7 ± (–7)2 – (4 × 1 × 8) x = 2 7 ± 49 – 32 Some pupils might find it easier to rewrite the formula with the coefficient 1 written before the x2 in examples such as this. This avoids the error of substituting 0 for a. Sometimes it is sufficient to leave the solution in surd form. At other times it may be necessary to give the answer to a given number of decimal places or significant figures. x = 2 7 + 17 or x = 2 7 – 17 x = 5.562 x = 1.438 (to 3 d.p.)

Using the quadratic formula Use the quadratic formula to solve 2x2 + 5x – 1 = 0. 2x2 + 5x – 1 = 0 x = –b ± b2 – 4ac 2a x = 2 × 2 –5 ± 52 – (4 × 2 × –1) x = 4 –5 ± 25 + 8 x = 4 –5 + 33 or x = 4 –5 – 33 x = 0.186 x = –2.686 (to 3 d.p.)

Using the quadratic formula Use the quadratic formula to solve 9x2 – 12x + 4 = 0. 9x2 – 12x + 4 = 0 x = –b ± b2 – 4ac 2a x = 2 × 9 12 ±  (–12)2 – (4 × 9 × 4) x = 18 12 ± 144 – 144 Point out that whenever b2 – 4ac is 0, there will only be one solution to the quadratic equation. x = 18 12 ± 0 There is only one solution, x = 2 3

Using the quadratic formula Use the quadratic formula to solve x2 + x + 3 = 0. 1x2 + 1x + 3 = 0 x = –b ± b2 – 4ac 2a x = 2 × 1 –1 ± 12 – (4 × 1 × 3) x = 2 –1 ± 1 – 12 Ask pupils to explain why we cannot find the square root of a negative number. Point out that whenever b2 – 4ac is negative, there will be no real solution to the quadratic equation. x = 2 –1 ± –11 We cannot find –11 and so there are no solutions.

Using b2 – 4ac From using the quadratic formula, –b ± b2 – 4ac x = 2a we can see that we can use the expression under the square root sign, b2 – 4ac, to decide how many solutions there are. When b2 – 4ac is positive, there are two solutions. When b2 – 4ac is equal to zero, there is one solution. When b2 – 4ac is negative, there are no solutions.

Using b2 – 4ac We can demonstrate each of these possibilities using graphs. Remember, if we plot the graph of y = ax2 + bx + c the solutions to the equation ax2 + bx + c = 0 are given by the points where the graph crosses the x-axis. y x b2 – 4ac is positive y x b2 – 4ac is zero y x b2 – 4ac is negative Two solutions One solution No solutions

Using b2 – 4ac

A6.5 Problems leading to quadratic equations Contents A6 Quadratic equations A A6.1 Solving quadratic equations by factorization A A6.2 Completing the square (Not Required) A A6.3 Using the quadratic formula A A6.4 Equations with fractions (Not Required) A A6.5 Problems leading to quadratic equations

Problems leading to quadratic equations Some real-life problems can be solved using quadratic equations. For example, Jenny drives 24 miles to get to work. On the way home she is caught in traffic and drives 20 miles per hour slower than on the way there. If her total journey time to work and back is 1 hour, what was her average speed on the way to work? Remember, time taken = distance average speed Let Jenny’s average speed on the way to work be x.

Problems leading to quadratic equations Jenny drives 24 miles to get to work. On the way home she is caught in traffic and drives 20 miles per hour slower than on the way there. If her total journey time to work and back is 1 hour, what was her average speed on the way to work? Jenny’s time taken to get to work = 24 x Jenny’s time taken to get home from work = 24 x – 20 Total time there and back = 24 x – 20 x + = 1 Solving this equation will give us the value of x, Jenny’s average speed on the way to work.

Problems leading to quadratic equations 24 x – 20 x + = 1 Start by multiplying through by x(x – 20) to remove the fractions: 24(x – 20) + 24x = x(x – 20) expand the brackets: 24x – 480 + 24x = x2 – 20x simplify: 48x – 480 = x2 – 20x collect terms on the r.h.s.: 0 = x2 – 68x + 480 Discuss the fact that although x = 8 solves the equation it cannot be the solution to this particular problem. If Jenny’s speed on the way to work was 8 miles per hour her speed on the way home from work would be negative (because it is x – 20 miles per hour). The solution that make sense is 60 miles per hour. factorize: 0 = (x – 60)(x – 8) We have two solutions x = 60 and x = 8. Which of these solutions is not possible in this situation?

Problems leading to quadratic equations The only solution that makes sense is x = 60 miles per hour. If Jenny’s average speed on the way to work was 8 miles per hour her average speed on the way home would be –12 miles per hour, a negative number. We can therefore ignore the second solution. When practical problems lead to quadratic equations it is very often the case that only one of the solution will make sense in the context of the original problem. This is usually because many physical quantities, such as length, can only be positive.

Problems leading to quadratic equations The lengths of the two shorter sides in a right-angled triangle are x cm and (x – 7) cm. If the length of the hypotenuse is (x + 1) cm, find the value of x and hence the lengths of all three sides of the triangle. Let’s start by drawing a diagram, x + 1 x – 7 x Ask pupils to recall Pythagoras’ Theorem in words or symbols. We can use Pythagoras’ Theorem to write an equation in terms of x.

Problems leading to quadratic equations The lengths of the two shorter sides in a right-angled triangle are x cm and (x – 7) cm. If the length of the hypotenuse is (x + 1) cm, find the value of x and hence the lengths of all three sides of the triangle. x2 + (x – 7)2 = (x + 1)2 x2 + (x – 7)(x – 7) = (x + 1)(x + 1) expand: x2 + x2 – 7x – 7x + 49 = x2 + x + x + 1 simplify: 2x2 – 14x + 49 = x2 + 2x + 1 Ask pupils to recall Pythagoras’ Theorem in words or symbols. collect on the l.h.s.: x2 – 16x + 48 = 0 factorize: (x – 4)(x – 12) = 0 x = 4 or x = 12

Problems leading to quadratic equations The lengths of the two shorter sides in a right-angled triangle are x cm and (x – 7) cm. If the length of the hypotenuse is (x + 1) cm, find the value of x and hence the lengths of all three sides of the triangle. If x = 4 then the lengths of the three sides are, 4 cm, 4 – 7 = –3 cm and 4 + 1 = 5 cm We cannot have a side of negative length and so x = 4 is not a valid solution. Check the final solution using Pythagoras’ Theorem. If x = 12 then the lengths of the three sides are, 12 cm, 12 – 7 = 5 cm and 12 + 1 = 13 cm So, the shorter sides are 12 cm and 5 cm and the hypotenuse is 13 cm.