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GCSE: Quadratic Inequalities

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1 GCSE: Quadratic Inequalities
Dr J Frost Last modified: 7th May 2017

2 Reminder: Inequalities
We want to solve 2π‘₯+1>5 How many solutions are there to this inequality? Infinitely many! π‘₯ could be 3, 8, 4.29, etc. Bro Side Notes: Equations and inequalities are not limited to just one solution: we have seen that quadratics equations often have 2 solutions, and in this example it would not be possible to β€˜list’ all the solutions because there are infinitely many! ? How do we represent the set of all solutions to this inequality? π‘₯>2 i.e. β€œany real number greater than 2” ? In this particular example, known as a linear inequality, the range of solutions was simple. But for quadratic inequalities, this might be more complex…

3 Quadratic Inequalities
By trial and error (or otherwise!), try and think of the range (or ranges) of values that satisfies the following inequality. Your values need not be whole numbers. π‘₯ 2 +2π‘₯βˆ’15>0 ? Anything above 3 or below -5 works, e.g. when π‘₯ is 4: βˆ’15=9 and 9>0 π‘₯<βˆ’5 π‘œπ‘Ÿ π‘₯>3 You may have spotted that the -5 and 3 are solutions to the equation π‘₯ 2 +2π‘₯βˆ’15=0. How then can we get the inequality from these values without guess work?

4 Click to Bro-Bolden >
Solving Quadratic Inequalities Solve π‘₯ 2 +2π‘₯βˆ’15>0 Step 1: Get 0 on one side (already done!) π‘₯+5 π‘₯βˆ’3 >0 ? Step 2: Factorise Step 3: Sketch and reason π‘₯ 𝑦 βˆ’5 3 𝑦=(π‘₯+5)(π‘₯βˆ’3) Since we sketched 𝑦=(π‘₯+5)(π‘₯βˆ’3) we’re interested where 𝑦>0, i.e. the parts of the line where the 𝑦 value is positive. Click to Bro-Bolden > What can you say about the π‘₯ values of points in this region? 𝒙<βˆ’πŸ“ What can you say about the π‘₯ values of points in this region? 𝒙>πŸ‘ ? π‘₯<βˆ’5 or π‘₯>3 Bro Note: If the 𝑦 value is β€˜strictly’ greater than 0, i.e. > 0, then the π‘₯ value is strictly less than -5. So the < vs ≀ must match the original question. ? ?

5 ? Sketch with highlighted region
Solving Quadratic Inequalities Solve π‘₯ 2 +2π‘₯βˆ’15≀0 Step 1: Get 0 on one side (already done!) π‘₯+5 π‘₯βˆ’3 ≀0 Step 2: Factorise Step 3: Sketch and reason π‘₯ 𝑦 βˆ’5 3 𝑦=(π‘₯+5)(π‘₯βˆ’3) ? Sketch with highlighted region ? Final solution βˆ’5≀π‘₯≀3 Again, what can we say about the π‘₯ value of any point in this region? Bro Note: We could write β€œπ‘₯β‰₯βˆ’5 and π‘₯≀3” but the above better expresses that π‘₯ is β€œbetween” -5 and 3. Bro Note: As discussed previously, we need ≀ rather than < to be consistent with the original inequality.

6 Further Examples Solve 𝒙 𝟐 +πŸ“π’™β‰₯βˆ’πŸ’ π‘₯ 2 +5π‘₯+4β‰₯0 π‘₯+4 π‘₯+1 β‰₯0
π‘₯ 2 +5π‘₯+4β‰₯0 π‘₯+4 π‘₯+1 β‰₯0 Solve 𝒙 𝟐 <πŸ— π‘₯ 2 βˆ’9<0 π‘₯+3 π‘₯βˆ’3 <0 ? ? π‘₯ 𝑦 βˆ’4 βˆ’1 𝑦=(π‘₯+4)(π‘₯+1) π‘₯ 𝑦 βˆ’3 3 𝑦=(π‘₯+3)(π‘₯βˆ’3) π‘₯β‰€βˆ’4 or π‘₯β‰₯βˆ’1 βˆ’3<π‘₯<3 Bro Note: The most common error I’ve seen students make with quadratic inequalities is to skip the β€˜sketch step’. Sod’s Law states that even though you have a 50% chance of getting it right without a sketch (presuming you’ve factorised correctly), you will get it wrong.

7 Test Your Understanding
Solve 𝒙 𝟐 βˆ’π’™βˆ’πŸ”β‰€πŸŽ π‘₯+2 π‘₯βˆ’3 ≀0 βˆ’2≀π‘₯≀3 1 ? Solve 𝟐 𝒙 𝟐 + π’™βˆ’πŸ 𝟐 >πŸ‘π’™ 2 π‘₯ 2 + π‘₯ 2 βˆ’4π‘₯+4>3π‘₯ 3 π‘₯ 2 βˆ’7π‘₯+4>0 3π‘₯βˆ’4 π‘₯βˆ’1 >0 π‘₯<1 π‘œπ‘Ÿ π‘₯> 4 3 2 ?

8 Exercises ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 1 Solve the following:
π‘₯ 2 βˆ’5π‘₯βˆ’6> 𝒙<βˆ’πŸ 𝒐𝒓 𝒙>πŸ” π‘₯ 2 +3π‘₯βˆ’4≀ βˆ’πŸ’β‰€π’™β‰€πŸ π‘₯ 2 βˆ’9β‰₯ π’™β‰€βˆ’πŸ‘ 𝒐𝒓 𝒙β‰₯πŸ‘ π‘₯ 2 +8π‘₯+12< βˆ’πŸ”<𝒙<βˆ’πŸ π‘₯ 2 βˆ’3π‘₯βˆ’10β‰₯ π’™β‰€βˆ’πŸ 𝒐𝒓 𝒙β‰₯πŸ“ π‘₯+3 2 ≀ βˆ’πŸ•β‰€π’™β‰€πŸ 4 π‘₯ 2 β‰₯ π’™β‰€βˆ’ πŸ‘ 𝟐 𝒐𝒓 𝒙β‰₯ πŸ‘ 𝟐 𝑦 2 +𝑦> π’š<βˆ’πŸ’ 𝒐𝒓 π’š>πŸ‘ 16βˆ’ π‘₯ 2 < 𝒙<βˆ’πŸ’ 𝒐𝒓 𝒙>πŸ’ π‘₯ π‘₯+2 2 < βˆ’πŸ“<𝒙<𝟐 10βˆ’ 2π‘₯+1 2 β‰₯π‘₯ βˆ’ πŸ— πŸ’ β‰€π’™β‰€πŸ (Requires knowledge of sketching cubics/quartics) π‘₯βˆ’1 π‘₯βˆ’2 π‘₯βˆ’3 > 𝟏<𝒙<𝟐 𝒐𝒓 𝒙>πŸ‘ π‘₯βˆ’1 π‘₯βˆ’2 π‘₯βˆ’3 < 𝒙<𝟏 𝒐𝒓 𝟐<𝒙<𝟐 π‘₯ 2 βˆ’9 π‘₯ 2 βˆ’4 > 𝒙<βˆ’πŸ‘ 𝒐𝒓 βˆ’πŸ<𝒙<𝟐 𝒐𝒓 𝒙>πŸ‘ a ? b ? 3 Solve the following: 2 π‘₯ 2 +3π‘₯+1≀ βˆ’πŸβ‰€π’™β‰€βˆ’ 𝟏 𝟐 3 π‘₯ 2 βˆ’7π‘₯βˆ’6> 𝒙<βˆ’ 𝟐 πŸ‘ 𝒐𝒓 𝒙>πŸ‘ 4 π‘₯ 2 βˆ’4π‘₯+1> 𝒙≠ 𝟏 𝟐 c ? ? a d ? ? b e ? c ? 2 a ? b ? c ? d ? e ? f ? N ? a b ? c ?


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