Solving quadratic equations

Using the = 0 method x( 3x + 2)(x -1) = 0 x = 0 3x + 2 = 0 and x – 1 = 0 3x = -2 x = 1 x = −2 3 Same thing for any number of brackets such as (x+2)(3x -4)( x + 1) = 0

Using the square root method Solve (2𝑥 −1) 2 = 16 square root both sides (2x – 1) = ± 16 remember to use the + or – sign (2x – 1) = ± 4 find x for +4 and then for -4 (2x – 1) = 4 and (2x – 1) = -4 2x = 5 2x = -3 x = 5 2 x = −3 2

By factorising 2 𝑥 2 +11𝑥+12=0 open factors of 2 𝑥 2 and 12 (2x + 3)(x + 4) = 0 2x 3 2x + 3 = 0 and x + 4 = 0 x 4 cross multiply to see if x = 3 2 and x = -4 they add up to 11

By using the equation 3 𝑥 2 - 8x + 2 = 0 where a = 3, b = -8 and c = 2 The equation is given in the equations page 𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎 𝑥= − −8 ± (−8) 2 −4(3)(2) 2 𝑥 3 use calculator to work it out twice, one for + and one for – x = 2.39 and x = 0.28

Problem involving quadratics Use the rectangle for find the value of x. x(2x -1) = 30 2𝑥 2 −𝑥=30 2𝑥 2 −𝑥 −30=0 Use formula x = −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎 to find the value of x Think : one value would not be suitable. Which one is it?

By completing the square Solve 𝑥 2 −6𝑥+8=0 take the 8 to the other side of the = 𝑥 2 −6 𝑥 = −8 take the number in front of x and divide it by 2 Square it and put it on either side of the = 𝑥 2 −6𝑥 + −3 2 = −8 + (−3) 2 On LHS, form a bracket with x (𝑥 ) 2 =1 What is in the bracket goes in the bracket! This (𝑥 −3) 2 = 1 means put the -3 in the bracket (x – 3) = ± 1 square root and solve.

Equations involving fractions 𝑥 −3 12 𝑥 −7 =70 𝑥 x 12 𝑥 - 7 x 𝑥 -3 x 12 𝑥 - 3 x -7 = 70 12 – 7 𝑥 - 36 𝑥 + 21 = 70 33 – 7 𝑥 - 36 𝑥 = 70 take the 70 on the other side and group like terms −37 −7𝑥 − 36 𝑥 =0 multiply by x all terms −7𝑥 2 −37𝑥 −36=0 now use the equation 𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎 𝑥= 37± (−37) 2 −4(−7)(−36) 2𝑥 −7 x = -4 or x = 9 7

Equations involving fractions 3 𝑥+2 + 1 𝑥+4 = 2 Use the lCM 3 𝑥+4 +(𝑥+2) (𝑥+2)(𝑥+4) = 2 simplify 4𝑥+14 (𝑥+2)(𝑥+4) = 2 cross multiply 4𝑥+14 = 2(𝑥+2)(𝑥+4) 4𝑥+14 = 2( 𝑥 2 +6𝑥+8) 4𝑥+14 =2𝑥 2 +12𝑥+16 0 = 2𝑥 2 +12𝑥 −4𝑥+16 −14 0= 2𝑥 2 +8𝑥+2 Solve by factorisation or using the equation.

3part problems Problems involving speed, distance and time. Problems involving cost, price per unit item and number of items Problems involving density, mass and volume. ( May 2013) These use equations which use an equation involving three values: 𝑠𝑝𝑒𝑒𝑑= 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑖𝑚𝑒 number of 𝑖𝑡𝑒𝑚𝑠= 𝑐𝑜𝑠𝑡 𝑢𝑛𝑖𝑡 𝑝𝑟𝑖𝑐𝑒

speed When its average speed is increased by 10km/h, the time taken for a car to make a journey of 105km is reduced by 15min. Find the average speed of the car. What do we need to find? The speed. Take this as x. What is the difference between the two situations? Time. Write down two equations involving time.

First journey : time= 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑠𝑝𝑒𝑒𝑑 𝑡𝑖𝑚𝑒= 105 𝑥 Second journey. The speed is now increased to x + 10 Second journey : time= 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑠𝑝𝑒𝑒𝑑 𝑡𝑖𝑚𝑒= 105 𝑥+10

The difference between the two journeys is that the second journey takes 15min less. Since the speed is in km/h, so units are in hours, we need to convert the 15min into hours. 15 60 = 1 4 ℎ Therefore the difference between the first and second journey is ¼ h The first journey takes more time since the speed is lower.

Time of first journey – time of second journey = ¼ h 105 𝑥 − 105 𝑥+10 = 1 4 Find LCM of the left hand side. 105 𝑥+10 −105𝑥 𝑥(𝑥+10) = 1 4 105𝑥+1050 −105𝑥 𝑥(𝑥+10) = 1 4 1050 𝑥(𝑥+10) = 1 4 cross multiply and solve