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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
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STARTER π₯ 2 β12π₯+36 π=π (1 distinct solution)
How many distinct real solutions do each of the following have? π₯ 2 β12π₯+36 π=π (1 distinct solution) π₯ 2 +π₯+3 No real solutions π₯ 2 β2π₯β1 π=πΒ± π (2 distinct solutions) ? ? ?
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π2 β 4ππ is known as the discriminant.
π π₯ 2 +ππ₯+π=0 π₯= βπΒ± π 2 β4ππ 2π Remember that a βrootsβ 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? π π βπππ<π Equal roots? π π βπππ=π Two distinct roots? π π βπππ>π ? ? ! ? π2 β 4ππ is known as the discriminant.
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The Discriminant π₯ 2 +3π₯+4 β7 π₯ 2 β4π₯+1 12 2 1 π₯ 2 β4π₯+4 2 π₯ 2 β6π₯β3
Equation Discriminant Number of Roots π₯ 2 +3π₯+4 ? β7 ? ? ? π₯ 2 β4π₯+1 12 2 ? ? 1 π₯ 2 β4π₯+4 ? ? 2 π₯ 2 β6π₯β3 60 2 ? ? π₯β4β3 π₯ 2 β47 ? ? 1β π₯ 2 4 2
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Problems involving the discriminant
? ? ? a) π=1, π=2π, π=3π+4 2π 2 β4 1 3π+4 =0 4 π 2 β12πβ16=0 π 2 β3πβ4=0 π+1 πβ4 =0 π=4 Bro Tip: Always start by writing out π, π and π explicitly. ? b) When π=4: π₯ 2 +8π₯+16=0 π₯+4 2 =0, π₯=β4 ?
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Test Your Understanding
π₯ 2 +5ππ₯+ 10π+5 =0 where π is a constant. Given that this equation has equal roots, determine the value of π. ? π=1, π=5π, π=10π+5 5π 2 β π+5 =0 25 π 2 β40πβ20=0 5 π 2 β8πβ4=0 5π+2 πβ2 =0 π=2
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Exercises Find the values of π for which π₯ 2 +ππ₯+9=0 has equal roots. π=Β±π Find the values of π for which π₯ 2 βππ₯+4=0 has equal roots. π=Β±π 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 + 1β3π π₯+ 5π+1 =0 has equal roots. Hence find the value of π₯ for this value of π. π=π, π=π Find the values of π for which π π₯ 2 + 2π+1 π₯=4 has equal roots. π=βπΒ± π π 1 ? 2 ? 3 ? 4 ? 5 ? Weβll revisit this topic after weβve done Inequalities.
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