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Quadratic Polynomial>
Applications of the Discriminant>
Applications of the Discriminant 2Applications of the Discriminant 2
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Question 1 of 6
1. Question
Using the discriminant, find the nature of the roots of the function:x2-6x-4=0- 1.
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2.
No real roots -
3.
Two real roots
Hint
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Nature of the Roots Discriminant (Δ) Two real roots Δ>0 One real root Δ=0 No real roots Δ<0 Discriminant Formula
Δ=b2−4acFirst, compute for the discriminantx2-6x-4=0a=1 b=-6 c=-4Δ = b2−4ac Discriminant Formula = (−6)2−4(1)(−4) Substitute values = 36+16 = 52 This is a positive value, which means Δ>0Therefore, the function has Two real rootsTwo real roots -
Question 2 of 6
2. Question
Identify which values of k will make the function below have no real rootskx2+4x+1=0-
1.
-4 <k<4 -
2.
k <4 -
3.
k >4 -
4.
k <0
Hint
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Nature of the Roots Discriminant (Δ) Two real roots Δ>0 One real root Δ=0 No real roots Δ<0 Discriminant Formula
Δ=b2−4acFirst, compute for the discriminantkx2+4x+1=0a=k b=4 c=1Δ = b2−4ac Discriminant Formula = 42−4(k)(1) Substitute values = 16-4k Remember that for a function to have one real root, Δ<0Substitute the Δ computed previously, and then solve for kΔ < 0 16-4k < 0 16-4k -16 < 0 -16 Subtract 16 from both sides -4k < -16 -4k÷-4 < -16÷-4 Divide both sides by -4 k > 4 An inequality flips when divided by a negative k>4 -
1.
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Question 3 of 6
3. Question
Identify which values of m will make the function below have real rootsx2+(m+2)x+1=0-
1.
-4 ≤m≤0 -
2.
m ≤0 and m≥4 -
3.
m ≤-4 and m≥0 -
4.
0 ≤m≤4
Hint
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Nature of the Roots Discriminant (Δ) Two real roots Δ>0 One real root Δ=0 No real roots Δ<0 Discriminant Formula
Δ=b2−4acFirst, compute for the discriminantx2+(m+2)x+1=0a=1 b=m+2 c=1Δ = b2−4ac Discriminant Formula = (m+2)2−4(1)(1) Substitute values = m2+4m+4-4 = m2+4m A function that has real roots can have either one or two real roots, hence Δ≥0Substitute the Δ computed previously, and then solve for mΔ ≥ 0 m2+4m ≥ 0 m(m+4) ≥ 0 m=0 m=-4 To determine which region around m=0 and m=-4 would be included, plot these points and make a rough sketch of m2+4mReplace the x axis with m axis and draw an upward parabola since 1 is positiveRemember that Δ must be positiveTherefore, m≤-4 and m≥0m≤-4 and m≥0 -
1.
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Question 4 of 6
4. Question
Identify which values of m will make the function below have one real root2x2+mx+m-2=0- m= (4)
Hint
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Chapters- Chapters
Nature of the Roots Discriminant (Δ) Two real roots Δ>0 One real root Δ=0 No real roots Δ<0 Discriminant Formula
Δ=b2−4acFirst, compute for the discriminant2x2+mx+m-2=0a=2 b=m c=m-2Δ = b2−4ac Discriminant Formula = m2−4(2)(m−2) Substitute values = m2-8m+16 Remember that the function must have one real root, hence Δ=0Substitute the Δ computed previously, and then solve for mΔ = 0 m2-8m+16 = 0 [insert cross method image with two m’s on the left and two -4’s on the right](m-4)(m-4) = 0 m=4 m=4 -
Question 5 of 6
5. Question
Identify which values of k will make the function below have no real rootskx2-4x+k=0-
1.
0 <k<2 -
2.
k <0 and k>2 -
3.
k <-2 and k>2 -
4.
-2 <k<2
Hint
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Chapters- Chapters
Nature of the Roots Discriminant (Δ) Two real roots Δ>0 One real root Δ=0 No real roots Δ<0 Discriminant Formula
Δ=b2−4acFirst, compute for the discriminantkx2-4x+k=0a=k b=-4 c=kΔ = b2−4ac Discriminant Formula = (−4)2−4(k)(k) Substitute values = 16-4k2 Remember that the function must have no real roots, hence Δ<0Substitute the Δ computed previously, and then solve for kΔ < 0 16-4k2 < 0 4(4-k2) < 0 4(2-k)(2+k) < 0 k=-2 k=2 To determine which region around k=-2 and k=2 would be included, plot these points and make a rough sketch of 16-4k2Replace the x axis with k axis and draw a downward parabola since -4 is negativeRemember that Δ must be negativeTherefore, k<-2 and k>2k<-2 and k>2 -
1.
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Question 6 of 6
6. Question
Identify which values of m will make the function below have two real roots2x2+mx+8=0-
1.
m <-8 and m>8 -
2.
0 <m<4 -
3.
m <-4 and m>8 -
4.
-8 <m<8
Hint
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Chapters- Chapters
Nature of the Roots Discriminant (Δ) Two real roots Δ>0 One real root Δ=0 No real roots Δ<0 Discriminant Formula
Δ=b2−4acFirst, compute for the discriminant2x2+mx+8=0a=2 b=m c=8Δ = b2−4ac Discriminant Formula = m2−4(2)(8) Substitute values = m2-64 Remember that the function must have two real roots, hence Δ>0Substitute the Δ computed previously, and then solve for mΔ > 0 m2-64 > 0 (m+8)(m-8) > 0 m=-8 m=8 To determine which region around m=-8 and k=8 would be included, plot these points and make a rough sketch of m2-64Replace the x axis with m axis and draw an upward parabola since 1 is positiveRemember that Δ must be positiveTherefore, m<-8 and m>8m<-8 and m>8 -
1.
Quizzes
- Sum & Product of Roots 1
- Sum & Product of Roots 2
- Sum & Product of Roots 3
- Sum & Product of Roots 4
- Solving Equations by Factoring 1
- Solving Equations Using the Quadratic Formula
- Completing the Square 1
- Completing the Square 2
- Intro to Quadratic Functions (Parabolas) 1
- Intro to Quadratic Functions (Parabolas) 2
- Intro to Quadratic Functions (Parabolas) 3
- Graph Quadratic Functions in Standard Form 1
- Graph Quadratic Functions in Standard Form 2
- Graph Quadratic Functions by Completing the Square
- Graph Quadratic Functions in Vertex Form
- Write a Quadratic Equation from the Graph
- Write a Quadratic Equation Given the Vertex and Another Point
- Quadratic Inequalities 1
- Quadratic Inequalities 2
- Quadratics Word Problems 1
- Quadratics Word Problems 2
- Quadratic Identities
- Graphing Quadratics Using the Discriminant
- Positive and Negative Definite
- Applications of the Discriminant 1
- Applications of the Discriminant 2
- Solving Reducible Equations