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Applications of the Discriminant>
Applications of the Discriminant 1Applications of the Discriminant 1
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Question 1 of 7
1. Question
Using the discriminant, find the nature of the roots of the function:5x2−6x+4=0- 1.
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2.
No real roots -
3.
One real root
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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 discriminant5x2−6x+4=0a=5 b=−6 c=4Δ = b2−4ac Discriminant Formula = (−6)2−4(5)(4) Substitute values = 36−80 = −44 This is a negative value, which means Δ<0Therefore, the function has No real rootsNo real roots -
Question 2 of 7
2. Question
Identify which values of k will make the function below have one real rootx2−(k−8)x+4=0-
1.
k=−4,2 -
2.
k=4,12 -
3.
k=−12,−4 -
4.
k=0
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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−(k−8)x+4=0a=1 b=k−8 c=4Δ = b2−4ac Discriminant Formula = (k−8)2−4(1)(4) Substitute values = k2−16k+64−16 = k2−16k+48 Remember that for a function to have one real root, Δ=0Substitute the Δ computed previously, and then solve for kΔ = 0 k2−16k+48 = 0 [insert cross method with two k’s on the right and −4 & −12 on the left](k−4)(k−12) = 0 k=4,12 k=4,12 -
1.
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Question 3 of 7
3. Question
Identify which values of k will make the function below have two real rootsx2−3kx+9=0-
1.
−2 <k<2 -
2.
−3 <k<3 -
3.
k <−3 and k>3 -
4.
k <−2 and k>2
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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 discriminantx2−3kx+9=0a=1 b=−3k c=9Δ = b2−4ac Discriminant Formula = (−3k)2−4(1)(9) Substitute values = 9k2−36 Remember that for a function to have two real roots, Δ>0Substitute the Δ computed previously, and then solve for kΔ > 0 9k2−36 > 0 9(k2−4) > 0 9(k−2)(k+2) > 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 9k2−36Replace the x axis with k axis and draw an upward parabola since 9 is positiveRemember that Δ must be positiveTherefore, k<−2 and k>2k<−2 and k>2 -
1.
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Question 4 of 7
4. Question
Identify which values of k will make the function below have real rootsx2+(k+2)x+4=0-
1.
−2 ≤k≤6 -
2.
k ≤−6 and k≥2 -
3.
k ≤−2 and k≥6 -
4.
−6 ≤k≤2
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+(k+2)x+4=0a=1 b=k+2 c=4Δ = b2−4ac Discriminant Formula = (k+2)2−4(1)(4) Substitute values = k2+4k+4−16 = k2+4k−12 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 kΔ ≥ 0 k2+4k−12 ≥ 0 (k+6)(k−2) ≥ 0 k=−6 k=2 To determine which region around k=−6 and k=2 would be included, plot these points and make a rough sketch of k2+4k−12Replace the x axis with k axis and draw an upward parabola since 1 is positiveRemember that Δ must be positiveTherefore, k≤−6 and k≥2k≤−6 and k≥2 -
1.
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Question 5 of 7
5. Question
Identify which values of k will make the function below have one real rootkx2−4x+k=0-
1.
k=−2 -
2.
k=−2,2 -
3.
k=0,2 -
4.
k=2
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+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 one real root, 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 k=−2,2 -
1.
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Question 6 of 7
6. Question
Identify which values of m will make the function below have one real root2x2+mx+8=0-
1.
m=−8,8 -
2.
m=−8 -
3.
m=0,8 -
4.
m=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 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 one real root, hence Δ=0Substitute the Δ computed previously, and then solve for mΔ = 0 m2−64 = 0 (m+8)(m−8) = 0 m=−8,8 m=−8,8 -
1.
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Question 7 of 7
7. Question
For what value of k will the line y=3x−k be tangent to the parabola of y=2x2−x+3- k= (-1)
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=b2−4acEquate the two functions and solve for k in such a way that the discriminant (Δ) will be 0First, equate the two functions to create a single equationy=2x2−x+3y=3x−k2x2−x+3 = 3x−k 2x2−x+3−3x+k = 0 Move all values to the left 2x2−4x+3+k = 0 Next, compute for the discriminant2x2−4x+3+ka=2 b=−4 c=3+kΔ = b2−4ac Discriminant Formula = (−4)2−4(2)(3+k) Substitute values = 16−24−8k = −8−8k Remember that the line must be a tangent to the parabola which means they meet at one point and that the previous equation must have one root, hence Δ=0Substitute the Δ computed previously, and then solve for kΔ = 0 −8−8k = 0 −8−8k +8 = 0 +8 Add 8 to both sides −8k = 8 −8k÷(−8) = 8÷(−8) Divide both sides by −8 k = −1 k=−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