## Math 9 Chapter 4 Lesson 4: Solution formula of quadratic equation

## 1. Summary of theory

**Solution formula of quadratic equation**

We have the general equation: \(ax^2+bx+c=0 (a\neq 0)\)

Moving the term c to the right side, we have: \(ax^2+bx=-c\)

Since \(a\neq 0\) should divide both sides by a, we have: \(x^2+\frac{b}{a}x=-\frac{c}{a}\)

Convert to constant equality: \(x^2+2.\frac{1}{2}\frac{b}{a}x+\frac{b^2}{4a}-\frac{b^2 }{4a}=-\frac{c}{a}\)

\(\Leftrightarrow \left ( x+\frac{b}{2a} \right )^2=\frac{b^2-4ac}{4a^2}\)

Set \(\Delta =b^2-4ac\)

We have the following conclusions:

With the equation \(ax^2+bx+c=0 (a\neq 0)\) and the differential \(\Delta =b^2-4ac\):

\(\Delta>0\) then the equation has 2 distinct solutions:

\(x_{1}=\frac{-b+\sqrt{\Delta }}{2a}\); \(x_{2}=\frac{-b-\sqrt{\Delta }}{2a}\)

\(\Delta=0\) then the equation has a double solution \(x=x_{1}=x_{2}=-\frac{b}{2a}\)

\(\Delta<0\) the equation has no solution.

## 2. Illustrated exercise

### 2.1. Basic exercises

**Question 1:** Fill in the blanks (…) with the appropriate expressions below:

a) If \(\Delta \) > 0, then from equation (2) it follows that \(x + \displaystyle{b \over {2a}} = \pm …\)

Therefore, equation (1) has two solutions x1 = …, x2 = …

b) If \(\Delta \) = 0 then from equation (2) it follows that \({\left( {x + \displaystyle{b \over {2a}}} \right)^2} = … \)

Therefore, equation (1) has a double solution x = …

**Solution guide**

a) If \(\Delta \) > 0 then from equation (2) it follows that \(x + \displaystyle{b \over {2a}} = \displaystyle \pm {{\sqrt \Delta } \over {2a }}\)

Thus, equation (1) has two solutions \({x_1} = \displaystyle{{ { – b + \sqrt \Delta } } \over {2a}};\,\,\,{x_2} = {{ { – b – \sqrt \Delta }} \over {2a}}\,\)

b) If \(\Delta \) = 0 then from equation (2) it follows that \({\left( {x + \displaystyle{b \over {2a}}} \right)^2} = 0\)

Therefore, equation (1) has a double solution \(x = \displaystyle{{ – b} \over {2a}}\)

**Verse 2: **Apply the solution formula to solve the equations:

a) \(5x^2 – x +2 = 0\)

b) \(4x^2 – 4x + 1 = 0\)

c) \(-3x^2+ x + 5 = 0\)

**Solution guide**

a) Consider the equation \(5x^2 – x +2 = 0\) where \(a = 5; b = -1; c = 2\)

\(\Delta = {b^2} – 4ac = {\left( { – 1} \right)^2} – 4.5.2 = 1 – 40 = – 39 < 0\)

So the above equation has no solution.

b) Consider the equation \(4x^2 – 4x + 1 = 0\) where \(a = 4; b = -4; c = 1\)

\(\Delta = {b^2} – 4ac = {\left( { – 4} \right)^2} – 4.4.1 = 16 – 16 = 0\)

\( \Rightarrow \) equation with double solution

\(\displaystyle x = {{ – b} \over {2a}} = {{ – \left( { – 4} \right)} \over {2.4}} = {1 \over 2}\)

So the equation has a solution \(\displaystyle x = {1 \over 2}\)

c) Consider the equation \(-3x^2 + x + 5 = 0\) where \(a = -3; b = 1; c = 5\)

\(\Delta = {b^2} – 4ac = {1^2} – 4.\left( { – 3} \right).5 = 1 + 60 =61> 0\)

Therefore \(\Delta \) > 0 should apply the solution formula, the equation has 2 distinct solutions

\(\displaystyle{x_1} = {{1 – \sqrt {61} } \over 6};\,\,{x_2} = {{1 + \sqrt {61} } \over 6}\)

### 2.2. Advanced exercises

**Question 1: **Explain why when \(\Delta < 0\) the equation has no solution.

**Solution guide**

Consider the equation \(\left( 2 \right)\)

\({\left( {x + \dfrac{b}{{2a}}} \right)^2} = \dfrac{{{b^2} – 4ac}}{{4{a^2}}} \) (Page 44 Textbook)

Or \({\left( {x + \dfrac{b}{{2a}}} \right)^2} = \dfrac{\Delta }{{4{a^2}}}\) (because \( \Delta=b^2-4ac\))

If \(\Delta < 0\) then \(\dfrac{\Delta }{{4{a^2}}} < 0\) where \({\left( {x + \dfrac{b}{{2a) }}} \right)^2} \ge 0\) for every \(x\) so the equation \({\left( {x + \dfrac{b}{{2a}}} \right)^2} = \dfrac{\Delta }{{4{a^2}}}\)no solution

The given equation \(a{x^2} + bx + c = 0\) has no solution.

**Verse 2: **Given the equation: \(-x^2+2x+2017^{2017}=0\). Without solving the equation, tell how many solutions the above equation has.

**Solution guide**

We have, \(\Delta =b^2-4ac\).

Notice \(b^2>0\); \(ac=-2017^{2017}<0\Rightarrow 4ac>0\)

So \(\Delta >0\forall x\epsilon \mathbb{R}\)

The equation has 2 distinct solutions.

## 3. Practice

### 3.1. Essay exercises

**Question 1:** Determine the coefficients \(a, b, c\) and then solve the equation:

a) \(2{x^2} – 2\sqrt 2 x + 1 = 0\)

b) \(\displaystyle 2{x^2} – \left( {1 – 2\sqrt 2 } \right)x – \sqrt 2 = 0\)

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

d) \(3{x^2} + 7.9x + 3.36 = 0\)

**Verse 2: **Solve the equation graphically.

Given the equation \(2{x^2} + x – 3 = 0\)

a) Draw the graphs of two functions: \(y = 2{x^2},y = – x + 3\) in the same coordinate plane.

b) Find the coordinates of each intersection of the two graphs. Explain why these coordinates are all solutions of the given equation.

Solve the given equation using the solution formula, compare with the results found in b.

**Question 3:** For each of the following equations, find the value of m so that the equation has a double solution:

a) \(m{x^2} – 2\left( {m – 1} \right)x + 2 = 0\)

b) \(3{x^2} + \left( {m + 1} \right)x + 4 = 0\)

**Question 4:** For each of the following equations, find the values of \(m\) for which the equation has a solution; Calculate the solution of the equation in terms of \(m\):

a) \(m{x^2} + \left( {2x – 1} \right)x + m + 2 = 0\)

b) \(2{x^2} – \left( {4m + 3} \right)x + 2{m^2} – 1 = 0\)

### 3.2. Multiple choice exercises

**Question 1:** The solution of the equation \(x^2-12x+36=0\) is:

A. \(6\)

B. \(-6\)

C. \(\pm 6\)

D. \(6\) and \(12\)

**Verse 2:** Let m be a parameter of the unknown quadratic equation x: \(x^2-2(m-1)x-3-m=0\) has a solution

A.For every m

B. \(m>0\)

C. \(m\leq 0\)

D. \(m\geq 0\)

**Question 3:** The solution of the equation \(x^2+100x+2500=0\) is:

A. \(50\)

B. \(-50\)

C. \(\pm 50\)

D. \(\pm 100\)

**Question 4:** Without solving the equation, how many solutions does the equation \(5x^2+9x-1=0\) have?

A. Inexperienced

B. 1 double root

C. 2 distinct solutions

D. Countless solutions

**Question 5:** For what value of m does the quadratic equation \(x^2+5x-m=0\) have exactly 1 solution?

A. \(\frac{25}{4}\)

B. \(-\frac{25}{4}\)

C. \(\frac{25}{2}\)

D. \(-\frac{25}{2}\)

## 4. Conclusion

Through this lesson, help students

- Remember the difference =b
^{2}−4ac. Under what conditions of D, the equation has no solution, has a double solution, and has two distinct solutions. - Apply the solution of quadratic equations to solve quadratic equations fluently.
- Write the difference = b
^{2}−4ac. It is possible to solve quadratic equations with one unknown by using differential numbers.

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