Unit 2: Topic 6: Application of Viete’s Theorem

Competencies

· State Viete’s theorem

· Apply Viete’s theorem to solve problems

Suggested ways of teaching this topic: Presentation of the teacher followed by individual consolidation and practice activities

Starter Activities

Before coming to the discussion on Viete’s theorem and its application, the students shall understand the nature of the roots of a quadratic equation. The teacher might start from “A Question and Answer Session” where students can be asked to solve the following quadratic equation and find out the relationship between the roots and the coefficients of the equation. The process goes as follows:

Example: Solve each of the following equations by using the quadratic formula:

a) 2x² – 3x – 2 = 0

b) 4x²– 4x + 1 = 0

c) 3x² – 2x + 1 = 0

Solution:

a) 2x² – 3x – 2 = 0

Þ a = 2, b = -3, c = -2

b) 4x²– 4x + 1 = 0

Þ a = 4, b = - 4, c = 1

= =

c) 3x² – 2x + 1 = 0

Þ a = 3, b = -2, c = 1

Since, where a> 0, is undefined; this equation has no solution.

At this stage, the teacher shall ask students to observe the value of b² - 4ac (the expression inside the square root)for each of the three quadratics. The expected results are:

For a); b² - 4ac = 25, which is greater than 0

For b); b² - 4ac = 0

For c); b² - 4ac = - 8, which is less than zero.

The next is asking an important question: “What do you observe from the sign of b² – 4ac and the number solutions of these equations?”

Expected Answer

· From the examples solved above, it is clear that the expression b² - 4ac is the key to determine the nature of roots of a given quadratic equation.

· The expression b² - 4ac is called the discriminate and its value will enable us to make predictions regarding the types of roots that we will obtain.

Lesson Notes

Let D = b² – 4ac.

1. If D > 0, then the equation ha 2 distinct solution

2. If D = 0, then the equation has exactly 1 solution

3. If D < 0, then the equation has No solution

Example:

Find the discriminate of the following equations and determine their nature of roots, without actually solving any of them:

a) x² + 2x – 5 = 0

b) 2x² – x + 1 = 0

c) x² + 5x – 6 = 0

d) x² + 4x +4 = 0

Solution:

a) x² + 2x – 5 = 0

Þ a= 1, b= 2, c = -5

Þ D = b² – 4ac = 2² – (4)(1)(-5) = 24

D > 0 Þ 2 distinct Solution

b) 2x² – x + 1 = 0

Þ a = 2, b = -1, c= 1

Þ D=b²–4ac=(-1)²(0) - (4)(2)(1) = -7

D < 0 Þ No solution

c) x² + 5x – 6 = 0

Þ a = 1, b = 5, c = -6

Þ D = b² – 4ac = 5² – 4(1)(-6) = 25 + 24 = 49

D > 0 Þ two distinct solutions

d) x² + 4x +4 = 0

Þ a = 1, b = 4, c = 4

Þ D = b² – 4ac = 4² – 4(1)(4)

= 16 – 16 = 0

D = 0 Þ Exactly 1 solution

Applications of the Nature of Roots

Example 1:

For which values of k will the equation 2x²+kx + 1 = 0 have equal roots?

Solution: a = 2, b = k and c = 1

Þ D = b² – 4ac = (k)² -4(2)(1)

= k² – 8

But D = 0 (equal roots)

Þ k² – 8 = 0

Þ k = ±

Example 2:

Find the value of k for which the quadratic equationx² - 8x + k =0 will have exactly one solution.

Solution: a = 1, b= -8, c = k

Þ D = b² – 4ac = (-8)² – 4(1)(k) = 64 – 4k

But D = 0 (which means one solution)

Þ 64 – 4k = 0

Þ 64 = 4k

Þ k = 16

Example 3:

Find the value(s) of k for which the quadratic equation 3x² -2kx + 3 = 0 will have exactly one solution.

Solution:

3x² – 2kx + 3 = 0

Þ a = 3, b = -2k, c = 3

Þ D = b² – 4ac = (-2k)² – 4(3)(3)

= 4k² – 36

But D = 0 Þ 4k² = 36

Þ k = ± 3

Now, before stating the theorem, the teacher could take one of the quadratic equations with two roots and ask students to

1. add the roots, then compare the sum with the value of for that equation

2. multiply the roots , then compare the product with the value of for that equation

If x² + 5x – 6 = 0 were taken the roots are r₁ = - 6 and r₂ = 1.

Thus, r₁ + r₂= - 6 + 1 = -5 = and r₁ r₂= - 6 1 = - 6 =

This is time the theorem that explains the existing special relationship between the roots of a quadratic equation and its coefficients can be stated.

Viete’s Theorem

Theorem: If the root of ax² + bx + c = 0, a≠0 are r₁ and r₂, then

Example 4: Check Viete’s theorem using2x²+ 3x – 5 = 0

Þ a = 2, b = 3, c = -5

Checking Viete’s Theorem:

Sum:

Product:

Example 5: Without computing the roots a and b of the equation 3x² + 2x + 6 = 0 find

1. a+ b

2. ab

Solution:

From 3x² + 2x + 6 = 0, a = 3, b = 2, c = 6

1) a + b=

2) ab =

Example 6: Without computing the roots a and b of the equation 2x² – 5x – 12 = 0, find

Solutions:

2x² – 5x – 12 = 0 Þ a = 2, b = -5 , c = -12

a) =

Practice Exercises

1. Find the discriminate for each of the listed equations and determine their nature of roots, without actually solving any of the equations:

a) 2x² – 5x + 3 = 0

b) 9x² + x + 1 = 0

c) 4x²+ 4x + 1 = 0

d) x² – 5x – 3 = 0

2. How many solutions does each equation have?

a) 3x² – x + 2 = 0

b) 2x – x² + 15 = 0

c) 1 – 2x + x² = 0

3. If one of the roots and the product of the roots of the equation 6x² + Mx + N = 0 respectively are ½ and -2/3 , what are the values of M and N?

4. For what value of k will (k+3)x² + (3k+1)x + 1 = 0 have only one root?

5. A quadratic equation has two unequal roots. If the difference between them is 1 and the difference of their squares is 2, what will be the equation?

For the following questions choose the best answer

1. Which one of the following equations has two distinct roots?