## Math 9 Chapter 4 Lesson 5: The reduced solution formula

## 1.Theory summary

### 1.1. Compact test formula

For the quadratic equation \(ax^2+bx+c=0(a\neq 0)\), in many cases if \(b=2b’ (b\vdots 2)\) is set, will the calculation Is the math simpler?

\(b=2b’ \Rightarrow \Delta =(2b’)^2-4ac=4b’^2-4ac=4(b’^2-ac)\)

We have: \(\Delta ‘=b’^2-ac\)

From this, we come to the following conclusions:

For the quadratic equations \(ax^2+bx+c=0(a\neq 0)\) and \(b=2b’\), \(\Delta ‘=b’^2-ac\) then :

If \(\Delta ‘>0\) then the equation has two distinct solutions

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

If \(\Delta ‘=0\) then the equation has a double solution \(x=\frac{-b’}{a}\)

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

### 1.2. Apply

Let’s take a look at a few examples:

**Example 1: **Solve the equation using the reduced solution: \(3x^2+10x+5=0\)

**Solution guide**

\(\Delta ‘=5^2-5.3=10>0\Rightarrow \sqrt{\Delta ‘}=\sqrt{10}\)

So \(x_{1}=\frac{-5+\sqrt{10}}{3}; x_{2}=\frac{-5-\sqrt{10}}{3}\)

**Example 2: **Solve the equation using the reduced solution: \(5x^2-6\sqrt{2}x+1=0\)

**Solution guide**

\(\Delta ‘=(3\sqrt{2})^2-5.1=13>0\Rightarrow \sqrt{\Delta ‘}=13\)

So \(x_{1}=\frac{3\sqrt{2}+\sqrt{13}}{5}; x_{2}=\frac{3\sqrt{2}-\sqrt{13}}{ 5}\)

## 2. Illustrated exercise

### 2.1. Basic exercises

**Question 1: **Solve the equation \(5{x^2} + 4x – 1 = 0\) by filling in the blanks:

\(a = …;\,b’ = …;c = …\); \(\Delta ‘ = …;\,\sqrt {\Delta ‘} = …\)

Solution of the equation \({x_1} = …;\,{x_2} = …\)

**Solution guide**

\(a = 5;\,b’ = 2;c = – 1\);

\(\Delta ‘ = {(b’)^2} – ac = {2^2} – 5.\left( { – 1} \right) = 9;\,\sqrt {\Delta ‘} = 3\ )

Solution of the equation \({x_1} = \dfrac{{ – b’ + \sqrt {\Delta ‘} }}{a} = \dfrac{{ – 2 + 3}}{5} = \dfrac{1} {5};\\{x_2}= \dfrac{{ – b’ – \sqrt {\Delta ‘} }}{a} = \dfrac{{ – 2 – 3}}{5} = – 1.\)

**Verse 2:** Determine \(a, b’, c\) and then use the reduced solution formula to solve the equations:

a) \(3x^2 + 8x + 4 = 0\)

b) \(7{x^2} – 6\sqrt 2 x + 2 = 0\)

**Solution guide**

a) Consider the equation \(3x^2 + 8x + 4 = 0\) where \(a = 3; b’ = 4; c = 4\)

\(\Delta ‘ = {\left( {b’} \right)^2} – ac = {4^2} – 3.4 = 4 >0\)\(\Rightarrow \sqrt {\Delta ‘} = 2\ )

The equation has two distinct solutions:

\(\displaystyle {x_1} = {{ – 4 + 2} \over 3} = {{ – 2} \over 3};\,\,{x_2} = {{ – 4 – 2} \over 3} = – 2\)

b) Consider the equation \(7{x^2} – 6\sqrt 2 x + 2 = 0\) where \(a = 7;\,\,b’ = – 3\sqrt 2 ;\,\,c = 2\)

\(\Delta ‘ = {\left( {b’} \right)^2} – ac = {\left( { – 3\sqrt 2 } \right)^2} – 7.2 = 4\)

Derive \(\sqrt {\Delta ‘} = 2\)

Therefore, the equation has two distinct solutions:

\({x_1} = \dfrac{{3\sqrt 2 + 2}}{7};{x_2} = \dfrac{{3\sqrt 2 – 2}}{7}\)

### 2.2. Advanced exercises

**Question 1:** Find the value of parameter m so that the equation \(x^2+2mx-m+4=0\) has a solution.

__ Solution guide:__ We calculate the discriminant \(\Delta ‘\) of the above equation:

\(\Delta ‘=m^2-m+4\)

For the above equation to have a solution, then \(\Delta ‘\geq 0\Leftrightarrow m^2-m+4=m^2-2.\frac{1}{2}m+\frac{1}{4}+3 ,75> 0\forall m\epsilon \mathbb{R}\)

So, the above equation always has 2 distinct solutions.

**Verse 2: **Find the value of parameter m so that the equation \(x^{2}-mx+m-1=0\) has exactly 1 unique solution

__ Solution guide:__ We calculate the discriminant \(\Delta\) of the above equation:

\(\Delta =(-m)^2-4m+4=m^2-4m+4=(m-2)^2\)

Let the equation have a unique solution \(\Leftrightarrow \Delta =0\Leftrightarrow m=2\)

So for \(m=2\) the above equation has a unique solution.

## 3. Practice

### 3.1. Essay exercises

**Question 1:** Determine a, b’, c in each equation, and then solve the equation using the reduced solution formula:

a) \(5{x^2} – 6x – 1 = 0\)

b) \( – 3{x^2} + 14x – 8 = 0\)

c) \(- 7{x^2} + 4x = 3\)

**Verse 2:** For what values of \(x\) are the two expressions equal:

a) \({x^2} + 2 + 2\sqrt 2 \) and \(2\left( {1 + \sqrt 2 } \right)x\)

b) \(\sqrt 3 {x^2} + 2x – 1\) and \(2\sqrt 3 x + 3\)

c) \( – 2\sqrt 2 x – 1\) and \(\sqrt 2 {x^2} + 2x + 3\)

d) \({x^2} – 2\sqrt 3 x – \sqrt 3 \) and \(2{x^2} + 2x + \sqrt 3 \)

**Question 3: **Approximate solution of the equation (round to second decimal):

a) \(16{x^2} – 8x + 1 = 0\)

b) \(6{x^2} – 10x – 1 = 0\)

c) \(5{x^2} + 24x + 9 = 0\)

d) \(16{x^2} – 10x + 1 = 0\)

**Question 4: **For what value of \(x\) the values of the two functions are equal:

a) \(\displaystyle y = {1 \over 3}{x^2}\) and \(y = 2x – 3\)

b) \(\displaystyle y = – {1 \over 2}{x^2}\) and \(y = x – 8\)?

### 3.2. Multiple choice exercises

**Question 1: **Given the parametric equation m: \(x^2-mx+4=0\). For what value of parameter m, the above equation has no solution:

A. \(-4

B. \(m<4\)

C. \(m=4\)

D. \(m>4\)

**Verse 2: **Given the hidden function x parameter m \(x^2-(2m+1)x+m^2+m-6=0\). The value of m for the function to have a solution is:

A. \(m=5\)

B. \(m\geq 5\)

C. \(m\epsilon \mathbb{R}\)

D. \(m\neq 5\)

**Question 3:** Without solving the equation, the number of solutions to the equation \(x^2+4x-2017=0\) is:

A. 1 solution

B. 2 solutions

C. inexperienced

D. infinite number of solutions

**Question 4:** Given the equation hidden x parameter m: \(mx^2+2mx+m+2=0\). For what value of m does the above equation have a solution?

A. \(m>0\)

B. \(m<0\)

C. \(m\geq 0\)

D. \(m\leq 0\)

**Question 5:** Given an implicit function x parameter m \(x^2-2(m+1)x+m^2-4m+5=0\). The value of m so that the function has no solution is:

A. \(m<\frac{2}{3}\)

B. \(m>\frac{2}{3}\)

C. \(m>\frac{3}{2}\)

D. \(m<\frac{3}{2}\)

## 4. Conclusion

This lesson helps students:

- See the benefits of the reduced test formula.
- Determine b when necessary and remember the formula
- Remember and apply well the reduced test formula. Moreover, know how to use this formula thoroughly in every possible case to make the calculation simpler.

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