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C1 Discriminants (for Year 11s)

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1 C1 Discriminants (for Year 11s)
Dr J Frost Objectives: Understand the conditions under which a quadratic equation has no, equal or distinct roots. Last modified: 23rd August 2015

2 STARTER π‘₯ 2 βˆ’12π‘₯+36 𝒙=πŸ” (1 distinct solution)
How many distinct real solutions do each of the following have? π‘₯ 2 βˆ’12π‘₯+36 𝒙=πŸ” (1 distinct solution) π‘₯ 2 +π‘₯+3 No real solutions π‘₯ 2 βˆ’2π‘₯βˆ’1 𝒙=𝟏± 𝟐 (2 distinct solutions) ? ? ?

3 𝑏2 – 4π‘Žπ‘ is known as the discriminant.
π‘Ž π‘₯ 2 +𝑏π‘₯+𝑐=0 π‘₯= βˆ’π‘Β± 𝑏 2 βˆ’4π‘Žπ‘ 2π‘Ž Remember that a β€˜roots’ of an expression are the values of π‘₯ which make it 0. Looking at this formula, when in general do you think we have: No roots? 𝒃 𝟐 βˆ’πŸ’π’‚π’„<𝟎 Equal roots? 𝒃 𝟐 βˆ’πŸ’π’‚π’„=𝟎 Two distinct roots? 𝒃 𝟐 βˆ’πŸ’π’‚π’„>𝟎 ? ? ! ? 𝑏2 – 4π‘Žπ‘ is known as the discriminant.

4 The Discriminant π‘₯ 2 +3π‘₯+4 βˆ’7 π‘₯ 2 βˆ’4π‘₯+1 12 2 1 π‘₯ 2 βˆ’4π‘₯+4 2 π‘₯ 2 βˆ’6π‘₯βˆ’3
Equation Discriminant Number of Roots π‘₯ 2 +3π‘₯+4 ? βˆ’7 ? ? ? π‘₯ 2 βˆ’4π‘₯+1 12 2 ? ? 1 π‘₯ 2 βˆ’4π‘₯+4 ? ? 2 π‘₯ 2 βˆ’6π‘₯βˆ’3 60 2 ? ? π‘₯βˆ’4βˆ’3 π‘₯ 2 βˆ’47 ? ? 1βˆ’ π‘₯ 2 4 2

5 Problems involving the discriminant
? ? ? a) π‘Ž=1, 𝑏=2𝑝, 𝑐=3𝑝+4 2𝑝 2 βˆ’4 1 3𝑝+4 =0 4 𝑝 2 βˆ’12π‘βˆ’16=0 𝑝 2 βˆ’3π‘βˆ’4=0 𝑝+1 π‘βˆ’4 =0 𝑝=4 Bro Tip: Always start by writing out π‘Ž, 𝑏 and 𝑐 explicitly. ? b) When 𝑝=4: π‘₯ 2 +8π‘₯+16=0 π‘₯+4 2 =0, π‘₯=βˆ’4 ?

6 Test Your Understanding
π‘₯ 2 +5π‘˜π‘₯+ 10π‘˜+5 =0 where π‘˜ is a constant. Given that this equation has equal roots, determine the value of π‘˜. ? π‘Ž=1, 𝑏=5π‘˜, 𝑐=10π‘˜+5 5π‘˜ 2 βˆ’ π‘˜+5 =0 25 π‘˜ 2 βˆ’40π‘˜βˆ’20=0 5 π‘˜ 2 βˆ’8π‘˜βˆ’4=0 5π‘˜+2 π‘˜βˆ’2 =0 π‘˜=2

7 Exercises Find the values of π‘˜ for which π‘₯ 2 +π‘˜π‘₯+9=0 has equal roots. π’Œ=Β±πŸ” Find the values of π‘˜ for which π‘₯ 2 βˆ’π‘˜π‘₯+4=0 has equal roots. π’Œ=Β±πŸ’ Find the values of π‘˜ for which π‘˜ π‘₯ 2 +8π‘₯+π‘˜=0 has equal roots. Find the positive value of π‘˜ for which π‘₯ 2 + 1βˆ’3π‘˜ π‘₯+ 5π‘˜+1 =0 has equal roots. Hence find the value of π‘₯ for this value of π‘˜. π’Œ=πŸ‘, 𝒙=πŸ’ Find the values of π‘˜ for which π‘˜ π‘₯ 2 + 2π‘˜+1 π‘₯=4 has equal roots. π’Œ=βˆ’πŸΒ± πŸ‘ 𝟐 1 ? 2 ? 3 ? 4 ? 5 ? We’ll revisit this topic after we’ve done Inequalities.


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