## Math 9 Chapter 4 Lesson 6: Vietnamese system and application

## 1. Summary of theory

### 1.1. Vietnamese system

Recalling the old lesson about the quadratic equation \(ax^2+bx+c=0(a\neq 0)\) has 2 distinct solutions

\(x_1=\frac{-b+\sqrt{\Delta }}{2a}; x_2=\frac{-b-\sqrt{\Delta }}{2a}\)

We have: \(x_1+x_2=\frac{-2b+\sqrt{\Delta }-\sqrt{\Delta }}{2a}=-\frac{b}{a}\)

\(x_1.x_2=\frac{b^2-\Delta }{4a^2}=\frac{4ac}{4a^2}=\frac{c}{a}\)

**Viet’s theorem**

If \(x_1;x_2\) are two solutions of the equation \(ax^2+bx+c=0 (a\neq 0)\) then:

\(x_1+x_2=-\frac{b}{a}\) and \(x_1.x_2=\frac{c}{a}\)

**generality**

- If the equation \(ax^2+bx+c=0 (a\neq 0)\) has \(a+b+c=0\) then the equation has a solution of \(x_1=1\) and a solution. the other is \(x_2=\frac{c}{a}\).
- If the equation \(ax^2+bx+c=0 (a\neq 0)\) has \(a-b+c=0\) then the equation has a solution of \(x_1=-1\) and the other solution is \(x_2=-\frac{c}{a}\).

### 1.2. Find two numbers given their sum and product

Find 2 numbers knowing their sum is S and their product is P. Assuming one number is x then the other is \(Sx\)

So their product is rewritten as: \(x(Sx)=P\Leftrightarrow x^2-Sx+P=0\)

Set \(\Delta =S^2-4P\)

## 2. Illustrated exercise

### 2.1. Basic exercises

**Question 1: **Calculate \(x_1+x_2;x_1.x_2\)

\(\displaystyle{x_1} = {{ – b + \sqrt \Delta } \over {2a}};\,{x_2} = {{ – b – \sqrt \Delta } \over {2a}}\)

**Solution guide**

\(\eqalign{& {x_1} + {x_2} = {{ – b + \sqrt \Delta } \over {2a}} + {{ – b – \sqrt \Delta } \over {2a}} = {{ – 2b} \over {2a}} = {{ – b} \over a} \cr & {x_1}. {x_2} = \left( {{{ – b + \sqrt \Delta } \over {2a}} } \right).\left( {{{ – b – \sqrt \Delta } \over {2a}}} \right) = {{ {(-b)^2} – \Delta } \over {4{a ^2}}} \cr & = {{{b^2} – \left( {{b^2} – 4ac} \right)} \over {4{a^2}}} = {{4ac} \ over {4{a^2}}} = {c \over a} \cr} \)

**Verse 2: **Given the equation \(2x^2 – 5x + 3 = 0.\)

a) Determine the coefficients \(a, b, c\) and then calculate \(a + b + c.\)

b) Show that \( x_1 = 1\) is a solution of the equation.

c) Use Viet’s theorem to find \(x_2.\)

**Solution guide**

a) The equation \(2x^2 – 5x + 3 = 0\) has the coefficients \(a = 2; b = -5; c = 3\)

\( \Rightarrow a + b + c = 2 – 5 + 3 = 0\)

b) Substituting \(x = 1\) into the equation we get:

\(2.1^2 – 5.1 + 3 = 0 \Leftrightarrow 0=0\) (always true)

So \(x_1 = 1\) is a solution of the equation

c) According to Viet’s theorem we have:

\(\displaystyle{x_1}. {x_2} = {c \over a} = {3 \over 2} \Rightarrow 1. {x_2} = {3 \over 2} \Rightarrow {x_2} = {3 \over 2 }\)

### 2.2. Advanced exercises

**Question 1: **Given the equation \(3x^2 +7x + 4 = 0.\)

a) Determine the coefficients \(a, b, c\) and then calculate \(a – b + c.\)

b) Show that \( x_1 = -1\) is a solution of the equation.

c) Use Viet’s theorem to find \(x_2.\)

**Solution guide**

a) The equation \(3x^2 +7x + 4 = 0\) has the coefficients \(a = 3; b = 7; c = 4\)

\( \Rightarrow a – b + c = 3 – 7 + 4 = 0\)

b) Substituting \(x = -1\) into the equation we get:

\(3.(-1)^2 +7.(-1) + 4 = 0 \Leftrightarrow 0=0\) (always true)

So \(x_1 = -1\) is a solution of the equation

c) According to Viet’s theorem we have:

\(\displaystyle{x_1}. {x_2} = {c \over a} = {4 \over 3} \Rightarrow (-1). {x_2} = {4 \over 3} \Rightarrow {x_2} = {- 4 \over 3}\)

**Verse 2:** Find two numbers whose sum is equal to \(1\) and their product is equal to \(5.\)

**Solution guide**

The two numbers to find are the two solutions of the equation \(x^2-x+5=0\)

We have \( \Delta=(-1)^2-4.1.5=-19<0\) so the equation

no solution.

## Therefore, there are no two numbers that satisfy the problem condition.

3. Practice

**3.1. Essay exercises **

Question 1:

Solve the equation and then test the Viet equation:

a) \(3{x^2} – 2x – 5 = 0\)

b) \(5{x^2} + 2x – 16 = 0\)

**c) \(\displaystyle {1 \over 3}{x^2} + 2x – {{16} \over 3} = 0\)** d) \(\displaystyle {1 \over 2}{x^2} – 3x + 2 = 0\)

Verse 2:

Without solving the equation, using the Viet equation, calculate the sum and product of the solutions of each equation:

a) \(2{x^2} – 7x + 2 = 0\)

b) \(2{x^2} + 9x + 7 = 0\)

**c) \(\left( {2 – \sqrt 3 } \right){x^2} + 4x + 2 + \sqrt 2 = 0\) **d) \(1,4{x^2} – 3x + 1,2 = 0\)

Question 3:

Calculating the solution of the equation:

a) \(7{x^2} – 9x + 2 = 0\)

b) \(23{x^2} – 9x – 32 = 0\)

**c) \(1975{x^2} + 4x – 1979 = 0\)** d) \(\left( {5 + \sqrt 2 } \right){x^2} + \left( {5 – \sqrt 2 } \right)x – 10 \)\(\,= 0\)

Question 4:

Use the Vietnamese relation to find the solution \(x_2\) of the equation and then find the value of \(m\) in each of the following cases:

a) The equation \({x^2} + mx – 35 = 0\), knows the solution \(x_1= 7\).

b) The equation \({x^2} – 13x + m = 0,\) knows the solution \(x_1 = 12,5\).

### c) The equation \(4{x^2} + 3x – {m^2} + 3m = 0,\) knows the solution \(x_1 = -2\).

**3.2. Multiple choice exercises **

Question 1:

Given a hidden equation x with parameter m: \(x^2-(2m+3)x+m^2-3=0\)

The value of m so that the equation has both negative solutions is:

A. \(-\frac{7}{4}\leq m\leq \sqrt{3}\)

B. \(-\frac{7}{4}\leq m<-\sqrt{3}\)

**C. \(-\frac{7}{4}\leq m\leq -\sqrt{3}\) **D. \(m\geq \frac{-7}{4}\)

Verse 2:

Given the equation \(-x^2+8x-17=0\). The sum and product of the two solutions of the above equation are:

A. \(S=8; P=17\)

B. \(S=-8; P=17\)

**C. \(S=8; P=-17\)** D. Can’t find

Question 3:

The sum and product of 2 solutions of the equation \(x^2+6x-2017=0\) are respectively:

A. \(S=-6;P=2017\)

B. \(S=6;P=-2017\)

**C. \(S=6;P=2017\)** D. \(S=-6;P=-2017\)

Question 4:

Write a quadratic equation, knowing that it has a double solution \(x=5\)

A. \(x^2+10x+25=0\)

B. \(x^2+10x-25=0\)

**C. \(x^2-10x+25=0\) **D. \(x^2-10x-25=0\)

Question 5:

Given a hidden quadratic equation x parameter m: \(x^2-(2m+1)x+m^2+m-6=0\)

The value of m so that the equation has 2 solutions with opposite signs is:

A. \(m>2\)

B. \(m>-3\)

C. \(m>2\) or \(m<-3\)

## D. \(-3

4. Conclusion

- Through this lesson, students need to:
- State the Viet equation. Know how to represent the sum of squares and cubes of two solutions through the coefficients of the equation.
- Apply the applications of the Viet equation to:
- Brainstorm solutions of quadratic equations in the cases a + b + c = 0; a – b + c = 0 or cases where the sum and product of two solutions are integers with not very large absolute values.

Find two numbers whose sum and product are known.

Quadratic equation one hidden

Math 9

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