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GCSE: Quadratic Inequalities
Dr J Frost Last modified: 7th May 2017
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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 βlistβ all the solutions because there are infinitely many! ? How do we represent the set of all solutions to this inequality? π₯>2 i.e. βany real number greater than 2β ? In this particular example, known as a linear inequality, the range of solutions was simple. But for quadratic inequalities, this might be more complexβ¦
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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₯β15>0 ? Anything above 3 or below -5 works, e.g. when π₯ is 4: β15=9 and 9>0 π₯<β5 ππ π₯>3 You may have spotted that the -5 and 3 are solutions to the equation π₯ 2 +2π₯β15=0. How then can we get the inequality from these values without guess work?
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Click to Bro-Bolden >
Solving Quadratic Inequalities Solve π₯ 2 +2π₯β15>0 Step 1: Get 0 on one side (already done!) π₯+5 π₯β3 >0 ? Step 2: Factorise Step 3: Sketch and reason π₯ π¦ β5 3 π¦=(π₯+5)(π₯β3) Since we sketched π¦=(π₯+5)(π₯β3) weβre interested where π¦>0, i.e. the parts of the line where the π¦ value is positive. Click to Bro-Bolden > What can you say about the π₯ values of points in this region? π<βπ What can you say about the π₯ values of points in this region? π>π ? π₯<β5 or π₯>3 Bro Note: If the π¦ value is βstrictlyβ greater than 0, i.e. > 0, then the π₯ value is strictly less than -5. So the < vs β€ must match the original question. ? ?
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? Sketch with highlighted region
Solving Quadratic Inequalities Solve π₯ 2 +2π₯β15β€0 Step 1: Get 0 on one side (already done!) π₯+5 π₯β3 β€0 Step 2: Factorise Step 3: Sketch and reason π₯ π¦ β5 3 π¦=(π₯+5)(π₯β3) ? Sketch with highlighted region ? Final solution β5β€π₯β€3 Again, what can we say about the π₯ value of any point in this region? Bro Note: We could write βπ₯β₯β5 and π₯β€3β but the above better expresses that π₯ is βbetweenβ -5 and 3. Bro Note: As discussed previously, we need β€ rather than < to be consistent with the original inequality.
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Further Examples Solve π π +ππβ₯βπ π₯ 2 +5π₯+4β₯0 π₯+4 π₯+1 β₯0
π₯ 2 +5π₯+4β₯0 π₯+4 π₯+1 β₯0 Solve π π <π π₯ 2 β9<0 π₯+3 π₯β3 <0 ? ? π₯ π¦ β4 β1 π¦=(π₯+4)(π₯+1) π₯ π¦ β3 3 π¦=(π₯+3)(π₯β3) π₯β€β4 or π₯β₯β1 β3<π₯<3 Bro Note: The most common error Iβve seen students make with quadratic inequalities is to skip the βsketch stepβ. Sodβs Law states that even though you have a 50% chance of getting it right without a sketch (presuming youβve factorised correctly), you will get it wrong.
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Test Your Understanding
Solve π π βπβπβ€π π₯+2 π₯β3 β€0 β2β€π₯β€3 1 ? Solve π π π + πβπ π >ππ 2 π₯ 2 + π₯ 2 β4π₯+4>3π₯ 3 π₯ 2 β7π₯+4>0 3π₯β4 π₯β1 >0 π₯<1 ππ π₯> 4 3 2 ?
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Exercises ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 1 Solve the following:
π₯ 2 β5π₯β6> π<βπ ππ π>π π₯ 2 +3π₯β4β€ βπβ€πβ€π π₯ 2 β9β₯ πβ€βπ ππ πβ₯π π₯ 2 +8π₯+12< βπ<π<βπ π₯ 2 β3π₯β10β₯ πβ€βπ ππ πβ₯π π₯+3 2 β€ βπβ€πβ€π 4 π₯ 2 β₯ πβ€β π π ππ πβ₯ π π π¦ 2 +π¦> π<βπ ππ π>π 16β π₯ 2 < π<βπ ππ π>π π₯ π₯+2 2 < βπ<π<π 10β 2π₯+1 2 β₯π₯ β π π β€πβ€π (Requires knowledge of sketching cubics/quartics) π₯β1 π₯β2 π₯β3 > π<π<π ππ π>π π₯β1 π₯β2 π₯β3 < π<π ππ π<π<π π₯ 2 β9 π₯ 2 β4 > π<βπ ππ βπ<π<π ππ π>π a ? b ? 3 Solve the following: 2 π₯ 2 +3π₯+1β€ βπβ€πβ€β π π 3 π₯ 2 β7π₯β6> π<β π π ππ π>π 4 π₯ 2 β4π₯+1> πβ π π c ? ? a d ? ? b e ? c ? 2 a ? b ? c ? d ? e ? f ? N ? a b ? c ?
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