Inequalities

Introduction You will be familiar with solving Inequalities from GCSE maths and C1 In this chapter you will see how to solve some more complicated Inequalities You will also see how to avoid making a very common error! You will see how to use diagrams to help identify the correct regions for a question

Teachings for Exercise 1A

Inequalities 5𝑥+8=23 5𝑥+8>23 5𝑥=15 5𝑥>15 𝑥=3 𝑥>3 1A You can manipulate Inequalities in order to solve them Remember that solving an Inequality is very similar to solving an equation: 5𝑥+8=23 5𝑥+8>23 Subtract 8 Subtract 8 5𝑥=15 5𝑥>15 Divide by 5 Divide by 5 𝑥=3 𝑥>3 So the value of x in this case is 3 So the value of x in this case is anything greater than 3 The steps are effectively the same. However, there is one special situation when solving Inequalities that you need to be aware of… 1A

Multiply by -1. This REVERSES the sign! Inequalities You can manipulate Inequalities in order to solve them Remember that solving an Inequality is very similar to solving an equation: 6−2𝑥=2 6−2𝑥<2 Subtract 6 Subtract 6 −2𝑥=−4 −2𝑥<−4 Divide by 2 Divide by 2 −𝑥=−4 −𝑥<−4 Multiply by -1 Multiply by -1. This REVERSES the sign! 𝑥=4 𝑥>4 If you multiply or divide by a negative in an Inequality, you must reverse the direction of the sign… (you can check by substituting values back into the first step if you like!) 1A

Inequalities 2𝑥 2 <𝑥+3 2𝑥 2 −𝑥−3<0 (2𝑥−3)(𝑥+1)<0 2𝑥 2 <𝑥+3 You can manipulate Inequalities in order to solve them Solve the Inequality below: 2𝑥 2 <𝑥+3 Subtract x and subtract 3 2𝑥 2 −𝑥−3<0 Factorise (2𝑥−3)(𝑥+1)<0 2𝑥 2 <𝑥+3 So the ‘critical values’ are x = 3/2 and x = -1  Now draw a sketch. Use the critical values and the fact this is a positive quadratic… y Consider the Inequality – we want the range of values where the graph is below 0 So therefore: x -1 3/2 −1<𝑥< 3 2 1A

Inequalities [ ] [ ] 𝑥 2 𝑥−2 <𝑥+1 𝑥≠2 1A 𝑥 2 𝑥−2 <𝑥+1 𝑥 2 𝑥−2 <𝑥+1 You can manipulate Inequalities in order to solve them Solve the Inequality below: You MUST be careful in this situation. The normal process would be to multiply each side by (x – 2) However, this could be negative, there is no way to know for sure at this stage What you can do is multiply by (x – 2)2, as this will definitely be positive (as it has been squared) Then you rearrange and solve as in the previous example… You will need to use the ‘clever factorisation’ technique from FP1! Multiply by (x – 2)2 𝑥 2 (𝑥−2) 2 𝑥−2 <(𝑥+1) (𝑥−2) 2 Cancel an (x – 2) on the left 𝑥 2 𝑥−2 <𝑥+1 𝑥≠2 𝑥 2 (𝑥−2)<(𝑥+1) (𝑥−2) 2 Rearrange terms to one side 𝑥 2 𝑥−2 −(𝑥+1) (𝑥−2) 2 <0 Take out (x – 2) as a factor [ ] 𝑥−2 𝑥 2 − (𝑥+1)(𝑥−2) <0 Multiply out the inner bracket [ ] 𝑥−2 𝑥 2 −( 𝑥 2 −𝑥−2) <0 Simplify 𝑥−2 𝑥+2 <0 So the critical values of x are 2 and -2  Now sketch a graph to help with solving the inequality 1A

Inequalities 𝑥 2 𝑥−2 <𝑥+1 𝑥≠2 −2<𝑥<2 1A 𝑥 2 𝑥−2 <𝑥+1 𝑥 2 𝑥−2 <𝑥+1 You can manipulate Inequalities in order to solve them Solve the Inequality below: You MUST be careful in this situation. The normal process would be to multiply each side by (x – 2) However, this could be negative, there is no way to know for sure at this stage What you can do is multiply by (x – 2)2, as this will definitely be positive (as it has been squared) Then you rearrange and solve as in the previous example… You will need to use the ‘clever factorisation’ technique from FP1! We have shown that this Inequality is equivalent 𝑥−2 𝑥+2 <0 𝑥 2 𝑥−2 <𝑥+1 𝑥≠2 Plot a graph The shape is a positive quadratic The x-intercepts are 2 and -2 We want the region below the x-axis (< 0) Write this as an Inequality -2 2 −2<𝑥<2 1A

Inequalities [ ] [ ] [ ] 𝑥 𝑥+1 ≤ 2 𝑥+3 𝑥≠−1,𝑥≠−3 1A You can manipulate Inequalities in order to solve them Solve the Inequality below: Sometimes you need to multiply by two different denominators in order to cancel them both! As before, they both need to be squared to ensure they aren’t negative… 𝑥 𝑥+1 ≤ 2 𝑥+3 Multiply by (x+1)2(x+3)2 𝑥 (𝑥+1) 2 (𝑥+3) 2 𝑥+1 ≤ 2 (𝑥+1) 2 (𝑥+3) 2 𝑥+3 Cancel terms where appropriate 𝑥 𝑥+1 ≤ 2 𝑥+3 𝑥≠−1,𝑥≠−3 𝑥(𝑥+1) 𝑥+3 2 ≤2 𝑥+1 2 (𝑥+3) Rearrange and set equal to 0 𝑥 𝑥+1 𝑥+3 2 −2 𝑥+1 2 (𝑥+3)≤0 ‘Clever Factorisation’ [ ] (𝑥+1)(𝑥+3) 𝑥 𝑥+3 −2(𝑥+1) ≤0 Multiply out terms [ ] (𝑥+1)(𝑥+3) 𝑥 2 +3𝑥−2𝑥−2 ≤0 Simplify [ ] (𝑥+1)(𝑥+3) 𝑥 2 +𝑥−2 ≤0 Factorise the expression in the squared bracket (𝑥+1)(𝑥+3) (𝑥+2)(𝑥−1) ≤0 So the critical values of x are -1, -3, -2 and 1  Now sketch a graph to help with solving the inequality! 1A

Inequalities 𝑥 𝑥+1 ≤ 2 𝑥+3 𝑥≠−1,𝑥≠−3 −3<𝑥<−2 𝑜𝑟 −1<𝑥<−1 1A You can manipulate Inequalities in order to solve them Solve the Inequality below: Sometimes you need to multiply by two different denominators in order to cancel them both! As before, they both need to be squared to ensure they aren’t negative… 𝑥 𝑥+1 ≤ 2 𝑥+3 We have shown that this Inequality is equivalent (𝑥+1)(𝑥+3) (𝑥+2)(𝑥−1) ≤0 𝑥 𝑥+1 ≤ 2 𝑥+3 𝑥≠−1,𝑥≠−3 Plot a graph The shape is a positive quartic (same basic shape as a quadratic – ‘U’, just with more changes of direction!) The x-intercepts are -1, -3, -2, 1 We want the region below the x-axis (< 0) Write this using Inequalities -2 -1 -3 1 −3<𝑥<−2 𝑜𝑟 −1<𝑥<−1 1A

Teachings for Exercise 1B

Vertical asymptote at x = -1/3 Inequalities 𝑦=4−𝑥 You can use graphs to help solve Inequalities If you are solving an Inequality you can also find answers by drawing graphs of each side and looking for the region(s) where one graph is above/beneath the other… On the same axes sketch the graphs of the curves with equations: Find the points of intersection of the two graphs Solve the following equation: 4 The sketch for this graph is simple Downward sloping graph x and y intercepts at (4,0) and (0,4) 4 𝑦= 7𝑥 3𝑥+1 This one is more difficult Subbing in x = 0 or y = 0 will yield the intercept (0,0) 𝑦= 7𝑥 3𝑥+1 𝑦= 7𝑥 3𝑥+1 𝑦=4−𝑥 An asymptote will be at the value for x that makes the denominator 0 (as this is not possible) 3𝑥−1=0 𝑥=− 1 3 Solve Vertical asymptote at x = -1/3 7𝑥 3𝑥+1 <4−𝑥 1B

Rearrange to write in terms of x Multiply out the bracket Inequalities 𝑦=4−𝑥 You can use graphs to help solve Inequalities If you are solving an Inequality you can also find answers by drawing graphs of each side and looking for the region(s) where one graph is above/beneath the other… On the same axes sketch the graphs of the curves with equations: Find the points of intersection of the two graphs Solve the following equation: 4 The sketch for this graph is simple Downward sloping graph x and y intercepts at (4,0) and (0,4) 4 𝑦= 7𝑥 3𝑥+1 This one is more difficult Subbing in x = 0 or y = 0 will yield the intercept (0,0) Vertical asymptote at x = -1/3 𝑦= 7𝑥 3𝑥+1 𝑦= 7𝑥 3𝑥+1 𝑦=4−𝑥 Rearrange to write in terms of x  Multiply by (3x + 1) 𝑦 3𝑥+1 =7𝑥 Multiply out the bracket 3𝑥𝑦+𝑦=7𝑥 Subtract 3xy 𝑦=7𝑥−3𝑥𝑦 Factorise 7𝑥 3𝑥+1 <4−𝑥 𝑦=𝑥(7−3𝑦) Divide by (7 – 3y) 𝑦 7−3𝑦 =𝑥 1B

Horizontal asymptote at y = 7/3 Inequalities 𝑦=4−𝑥 You can use graphs to help solve Inequalities If you are solving an Inequality you can also find answers by drawing graphs of each side and looking for the region(s) where one graph is above/beneath the other… On the same axes sketch the graphs of the curves with equations: Find the points of intersection of the two graphs Solve the following equation: 4 The sketch for this graph is simple Downward sloping graph x and y intercepts at (4,0) and (0,4) 4 𝑦= 7𝑥 3𝑥+1 This one is more difficult Subbing in x = 0 or y = 0 will yield the intercept (0,0) Vertical asymptote at x = -1/3 𝑦= 7𝑥 3𝑥+1 𝑦 7−3𝑦 =𝑥 𝑦=4−𝑥 Find the value that would make the denominator 0 (which isn’t possible) 7−3𝑦=0 Add 3y 7=3𝑦 Divide by 3 7 3 =𝑦 7𝑥 3𝑥+1 <4−𝑥 Horizontal asymptote at y = 7/3 1B

Inequalities -1/3 𝑦=4−𝑥 You can use graphs to help solve Inequalities If you are solving an Inequality you can also find answers by drawing graphs of each side and looking for the region(s) where one graph is above/beneath the other… On the same axes sketch the graphs of the curves with equations: Find the points of intersection of the two graphs Solve the following equation: 4 The sketch for this graph is simple Downward sloping graph x and y intercepts at (4,0) and (0,4) 7/3 (0,0) 4 𝑦= 7𝑥 3𝑥+1 This one is more difficult Subbing in x = 0 or y = 0 will yield the intercept (0,0) Vertical asymptote at x = -1/3 Horizontal asymptote at y = 7/3 𝑦= 7𝑥 3𝑥+1 𝑦=4−𝑥 7𝑥 3𝑥+1 <4−𝑥 1B

Inequalities 1B -1/3 You can use graphs to help solve Inequalities 7/3 If you are solving an Inequality you can also find answers by drawing graphs of each side and looking for the region(s) where one graph is above/beneath the other… On the same axes sketch the graphs of the curves with equations: Find the points of intersection of the two graphs Solve the following equation: 𝑦= 7𝑥 3𝑥+1 4 7/3 (0,0) 4 b) The points of intersection will be where the two equations are set equal to each other 𝑦=4−𝑥 7𝑥 3𝑥+1 =4−𝑥 Multiply by (3x + 1) 𝑦= 7𝑥 3𝑥+1 7𝑥=(4−𝑥)(3𝑥+1) 𝑦=4−𝑥 Expand brackets 7𝑥=−3 𝑥 2 +11𝑥+4 Rearrange and set equal to 0 3 𝑥 2 −4𝑥−4=0 Factorise 3𝑥+2 𝑥−2 =0 Now you know the intersections 𝑥=− 2 3 𝑜𝑟 𝑥=2 7𝑥 3𝑥+1 <4−𝑥 1B

Inequalities 1B -1/3 You can use graphs to help solve Inequalities -2/3 The blue line is below the red line for x-values below -2/3 You can use graphs to help solve Inequalities If you are solving an Inequality you can also find answers by drawing graphs of each side and looking for the region(s) where one graph is above/beneath the other… On the same axes sketch the graphs of the curves with equations: Find the points of intersection of the two graphs Solve the following equation: 4 2 7/3 (0,0) 4 The blue line is below the red line for x-values between -1/3 and 2 7𝑥 3𝑥+1 <4−𝑥 𝑦= 7𝑥 3𝑥+1 𝑦=4−𝑥 Consider the colours (in this case) 7𝑥 3𝑥+1 <4−𝑥 So we want to know where the blue line is below the red line… 𝑥<− 2 3 − 1 3 <𝑥<2 7𝑥 3𝑥+1 <4−𝑥 1B

Inequalities 𝑥 2 −4𝑥 <3 𝑥 2 −4𝑥=0 𝑥(𝑥−4)=0 𝑥=0 𝑜𝑟 4 1B y =│x2 – 4x│ You can use graphs to help solve Inequalities Solve the Inequality: Start by sketching a graph of each side Remember for the modulus side, think about what the graph would look like without the modulus part… So the lowest value will be when x = 2 (so the minimum point will have a value of -4) This is important as when we reflect the lower part for the modulus, the peak will be above the y = 3 line We can see visually where the modulus graph is below y = 3, but we need the critical points… The original red line has equation y = x2 – 4x The reflected part has equation y = -(x2 – 4x) (2,4) 3 𝑥 2 −4𝑥 <3 2-√7 4 2+√7 (2,-4) 𝑥 2 −4𝑥=0 Intersection of y = 3 on the original graph 𝑥(𝑥−4)=0 𝑥 2 −4𝑥=3 Use completing the square (or the quadratic formula – this won’t factorise nicely!) 𝑥=0 𝑜𝑟 4 (𝑥−2) 2 − 4=3 Add 4 (𝑥−2) 2 =7 Square root 𝑥−2=± 7 Add 2 𝑥=2± 7 1B

Inequalities 𝑥 2 −4𝑥 <3 𝑥 2 −4𝑥=0 𝑥(𝑥−4)=0 𝑥=0 𝑜𝑟 4 1B y =│x2 – 4x│ You can use graphs to help solve Inequalities Solve the Inequality: Start by sketching a graph of each side Remember for the modulus side, think about what the graph would look like without the modulus part… So the lowest value will be when x = 2 (so the minimum point will have a value of -4) This is important as when we reflect the lower part for the modulus, the peak will be above the y = 3 line We can see visually where the modulus graph is below y = 3, but we need the critical points… The original red line has equation y = x2 – 4x The reflected part has equation y = -(x2 – 4x) (2,4) 3 𝑥 2 −4𝑥 <3 2-√7 1 3 2+√7 𝑥 2 −4𝑥=0 Intersection of y = 3 on the reflected graph 𝑥(𝑥−4)=0 −( 𝑥 2 −4𝑥)=3 ‘Expand’ the bracket 𝑥=0 𝑜𝑟 4 − 𝑥 2 +4𝑥=3 Rearrange and set equal to 0 𝑥 2 −4𝑥+3=0 Factorise 𝑥−3 𝑥−1 =0 𝑥=1 𝑜𝑟 𝑥=3 1B

Inequalities 𝑥 2 −4𝑥 <3 𝑥 2 −4𝑥 <3 𝑥 2 −4𝑥=0 𝑥(𝑥−4)=0 y =│x2 – 4x│ You can use graphs to help solve Inequalities Solve the Inequality: Start by sketching a graph of each side Remember for the modulus side, think about what the graph would look like without the modulus part… So the lowest value will be when x = 2 (so the minimum point will have a value of -4) This is important as when we reflect the lower part for the modulus, the peak will be above the y = 3 line We can see visually where the modulus graph is below y = 3, but we need the critical points… The original red line has equation y = x2 – 4x The reflected part has equation y = -(x2 – 4x) (2,4) 3 𝑥 2 −4𝑥 <3 2-√7 1 3 2+√7 𝑥 2 −4𝑥 <3 𝑥 2 −4𝑥=0 We need the ranges where the red graph is below the blue graph 𝑥(𝑥−4)=0 𝑥 2 −4𝑥 <3 𝑥=0 𝑜𝑟 4 2− 7 <𝑥<1 𝑜𝑟 3<𝑥<2+ 7 1B

Inequalities 3𝑥 +𝑥≤2 3𝑥=2−𝑥 −(3𝑥)=2−𝑥 3𝑥 ≤2−𝑥 4𝑥=2 −2𝑥=2 𝑥=0.5 𝑥=−1 1B y =│3x│ y = 3x You can use graphs to help solve Inequalities Sometimes rearranging the equation can make the sketch far easier to draw! Remember to be wary of whether you might by multiplying or dividing by a negative though! Solve: Now it is easier to sketch them both! Find the critical values. Remember to use y = 3x for the original red graph and y = -(3x) for the reflected part… 2 -1 0.5 2 y = 2 - x Intersection on the original red line Intersection on the reflected red line 3𝑥 +𝑥≤2 3𝑥=2−𝑥 −(3𝑥)=2−𝑥 Add x Add x 3𝑥 ≤2−𝑥 4𝑥=2 −2𝑥=2 Solve Solve 𝑥=0.5 𝑥=−1 1B

Inequalities 3𝑥 ≤2−𝑥 3𝑥 ≤2−𝑥 3𝑥 +𝑥≤2 −1≤𝑥≤0.5 3𝑥 ≤2−𝑥 1B y =│3x│ You can use graphs to help solve Inequalities Sometimes rearranging the equation can make the sketch far easier to draw! Remember to be wary of whether you might by multiplying or dividing by a negative though! Solve: Now it is easier to sketch them both! 2 -1 0.5 2 y = 2 - x 3𝑥 ≤2−𝑥 We want where the red line is below the blue line 3𝑥 ≤2−𝑥 3𝑥 +𝑥≤2 −1≤𝑥≤0.5 3𝑥 ≤2−𝑥 1B

Summary We have seen how to solve more complicated Inequalities We have seen how to avoid multiplying or dividing by a negative We have also seen how to use graphs to help answer questions!