# Chứng Minh Rằng: X^2 + X + 1 >0, Với Mọi X

Today you are going khổng lồ see 3 methods lớn **solve quadratic equations** that you need lớn know.

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**And these methods actually work. **

**quadratic equations**are.

Quadratic equations definitionHow lớn solve quadratic equationsQuadratic equations of the khung x² = k#1 – Factoring the quadratic và then applying the Null Factor law#2 – Completing the square #3 – Using quadratic equation formula

## Quadratic equations definition

Equations of the form **ax² + bx +c = 0 **where **a ≠ 0** are called **quadratic equations. **

They may have two, one, or zero **solutions**.

Here are some simple equations which clearly show the truth of this statement.

### Quadratic equation examples

**a) x² − 2x + 1 = 0** (in standard form)

a = 1 ≠ 0 , b = − 2 & c = 1.

**Solutions:** x = 1 (one solution)

**b) x² − 1 = 0**

a = 1 ≠ 0, b = 0 and c = − 1.

**Solutions:** x = 1 or x = − 1. (two solutions)

**c) x² + 1 = 0** means x² + 0x + 1 = 0

**Solutions:** None as x² is always ≥ 0. (zero)

But, **how vì chưng we find these solutions** without using trial & error?

In this lesson, we will discuss several methods for solving quadratic equations, & apply them to lớn practical problems.

## How lớn solve quadratic equations

### Quadratic equations of the khung x² = k

Consider the equation x² = 4.

Now 2 × 2 = 4, so x = 2 is one solution,

and (− 2) × (− 2) = 4, so x = − 2 is also a solution.

Thus, if x² = 4, then x = ±2 (±2 is read as ‘plus or minus 2’)

Solution of x² = kThis principle can be extended to other perfect squares.

For example, if** (x − 2)² = k** then **x − 2 = ±√k** provided **k > 0.**

**Examples**

**Example 1: Solve for x: **

**a) ****x² + 2 = 7 **

therefore** x² = 5 **(subtracting 2 from both sides)

∴ **x = ±√5 ** (±√5 is read as ‘plus or minus the square root of 5’)

**b) 3 − 2x² = 7 **

∴ **− 2x² = 4 **(subtracting 3 from both sides)

∴ **x² = − 2 **(dividing both sides by − 2)

which has no solutions as **x² cannot be **

**Example 2: Solve for x: **

**a) (x − 2)² = 25 (we vì not expand the LHS)**

**∴ x − 2 = ±√25**

**∴ x − 2 = ±5**

**Case 1: x − 2 = 5 **

**∴ x = 7.**

**Case 2: x − 2 = − 5 **

**∴ x = − 3.**

**b) (x + 1)² = 6**

** ∴ x + 1 = ±√6 **

**Case 1: x + 1 = √6 **

** ∴ x = √6 − 1**

**Case 2: x + 1 = − √6**

** ∴ x = − √6 − 1**

**For quadratic equations which are not of the form x² = k, we need an alternative method solution. One method is to factorize the quadratic và then apply the Null Factor law.**

**#1 – Factoring the quadratic và then applying the Null Factor law**

**The Null Factor law states that:**

**When the hàng hóa of two (or more) numbers is zero, then at least one of them must be zero.**

**So, if a × b = 0 then a = 0 or b = 0.**

**Steps for solving quadratic equations**

To use the Null Factor law when solving equations, we must have one side of the equation equal lớn zero.

If necessary, rearrange the equation so one side is**zero**.**Fully factorize**the other side (usually the LHS)Use the**Null Factor**law:**if a × b = 0 then a = 0 or b = 0.****Solve**the resulting linear equations.**Check**at least one of your solutions.Examples**Example 1: Solve for x: x² = 6x**

We rearrange the equation: **x² − 6x = 0**

We take out any common factors: **x(x − 6) = 0 **

We can use the **Null Factor** law: **x = 0** **or** **x − 6 = 0**

**Therefore, x = 0 or x = 6.**

**Example 2: Solve for x: x² + 2x = 8**

We rearrange the equation: **x² + 2x − 8 = 0**

We split the x-term ax² + bx + c, a ≠ 0.

find ac = 1 × (− 8) = − 8find the factors of ac which showroom to b: b = 2 so, these factors are 4 và − 2 (sum = 2 & product = − 8)replace 2x by 4x − 2xcomplete the factorizationIn this equation, we have

**x² + 2x − 8 = 0 **

⇒ **x² + 4x − 2x − 8 = 0**

⇒ **x(x + 4) − 2(x + 4) = 0**

⇒ **(x + 4)(x − 2) = 0**

We can use the **Null Factor** law:

**(x + 4)(x − 2) = 0 **

Case 1: **x + 4 = 0 ⇒ x = − 4.**

Case 2:** x − 2 = 0 ⇒ x = 2.**

**Check: If x = − 4 then (− 4)² +2(− 4) = 16 − 8 = 8 **

** If x = 2 then 2² +2(2) = 4 + 4 = 8 **

**So, x = 2 or − 4.**

**Example 3: Solve for x: 2x² = 3x − 1**

We rearrange the equation: **2x² − 3x + 1 = 0**

We have ac = 2, b = − 3 therefore, sum = − 3 and product = 2, the numbers are − 2 và − 1.

Therefore, we have **2x² − 2x − x + 1 = 0**

We factorize the pairs: **2x(x − 1) − (x − 1) = 0**

(x − 1) is a common factor: **(x − 1)( 2x − 1) = 0**

We can use the **Null Factor** law:

**(x − 1)( 2x − 1) = 0 ⇒ x − 1 = or 2x − 1 = 0**

**⇒ x = 1 or x = 1/2. **

**Example 4: Solve for x: **

We have **2(x − 2) = x(6 + x) (eliminating the algebraic fractions)**

**⇒ ** **2x − 4 = 6x + x² (expanding backets)**

**⇒ x² + 6x − 2x + 4 = 0 (and then making one side of the equation zero)**

**⇒ x² + 4x + 4 = 0 (recognise type: Perfect square)**

**⇒ (x + 2)² = 0**

**⇒ x + 2 = 0**

**⇒ x = − 2.Xem thêm: Lợi Ích Của Việc Sử Dụng Điện Thoại Trong Học Tập, Lợi Ích Của Việc Sử Dụng Điện Thoại Trong Lớp**

**Check: If x = − 2 then LHS = 2 & LRS = 2 **

### #2 – Completing the square

Some quadratic equations such as **x² + 6x + 2 = 0** cannot be solved by the methods above. This is because these quadratics have irrational solutions.

We, therefore, use a new technique where we **complete a perfect square**.

**What vì chưng we địa chỉ cửa hàng on lớn make a perfect square?**

**Halve** the coefficent of x. **Add the square of this number** khổng lồ both sides of the equation.

Consider **x² + 6x + 2 = 0.**

The first step is to keep the terms containing x on the LHS và write the constant term on the RHS. We get **x² + 6x = − 2.**

The coefficient of x is 6, so half this number is 3. We địa chỉ 3² or 9 khổng lồ both sides of the equation.

So, **x² + 6x + 9 = − 2 + 9 **

Therefore, **(x + 3)² = 7**

**⇒ x + 3 = ±√7**

**⇒ x = − 3 ±√7 **

**Example 1: Solve for x by completing the square:**

**a) x² + 4x – 4 = 0**

We move constant term khổng lồ RHS: **x² + 4x = 4**

We add (4/2)² = 2² to both sides: **x² + 4x + 2² = 4 + 2²**

We factorise LHS, simplify RHS: **(x + 2)² = 8**

**⇒ x + 2 = ± √8**

**⇒ x = − 2 ± √8 = − 2 ± 2√2 **

**b)**** x² − 2x + 7 = 0**

We move constant term to the RHS: **x² − 2x = − 7**

We địa chỉ cửa hàng (-2/2)² khổng lồ both sides: **x² − 2x + 1² = − 7 + 1² **

We factorise LHS, simplify RHS: **(x − 1)² = − 6**

which is impossible as no perfect square can be negative. Therefore no real solutions exist.

### #3 – Using quadratic equation formula

Many quadratic equations cannot be solved by factorization, & completing the square is rather tedious.

Consequently, the **quadratic formula** has been developed.

We just plug in the values of a, b và c, & do the calculations.

The ± means there are two solutions:

Here is an example with two answers:

**Example 1: Use the quadratic formula to solve for x: **

**a) x² − 2x − 2 = 0 **

We have a = 1, b = − 2, c = − 2. We plug in the values of a, b và c, and do the calculations.

Therefore,

**b) 2x² + 3x − 4 = 0**

We have a = 2, b = 3 & c = −4. We plug in the values of a, b và c, và do the calculations.

**But it’s not always worked out lượt thích that.**Consider **x² + 2x + 5 = 0.**

Using the quadratic formula, the solutions are:

However, in the real number system, **√-16** does not exist. Therefore, we say that **x² + 2x + 5 = 0 has no real solutions.**

If we graph **y = x² + 2x + 5 **we get:

The graph does not cross the x-axis, & this further justifies the fact that **x² + 2x + 5 = 0 has no real solutions.**

You can see **b² − 4ac **under the square root in the quadratic formula above. It is called the **discriminant. **The quadratic equation formula can be written as

**Notice that:**

**Δ = 0**,**x = −b/(2a)**is the**only solution**và is known as a**repeated root**.If**Δ >****0, √Δ**is a real number and so there are**two distinct real roots**If**Δ If a, b & c are rational &**

**Δ is a perfect square**then the quadratic equation has**two rational roots**which can be found by**factorization.****Example 2: Show that the following quadratic equation has no real solutions:**** 2x² − 3x + 4 = 0**

We have **Δ = b² − 4ac = (-3)² − 4×2×4 = −23 which is **

**Example 3:** **Given that kx² + 6x − 3 = 0 has a repeated root. Find k.**

**The discriminant Δ = b² − 4ac = 6² − 4×k×(−3) = 36 + 12k.**

**A quadratic equation has a repeated root when Δ = 0. **

**Therefore, 36 + 12k = 0.**

**⇒ 12k = − 36**

**⇒ k = −3. **

**Example 4: Use the quadratic formula khổng lồ solve:**

**We write the quadratic equation in standard form and plug in the values of a, b, & c, then bởi the calculations.**

**Example 5: Use the quadratic equation formula khổng lồ solve for x:**

**a. (x + 2)(x − 1) = 5**

**⇒ x² − x + 2x − 2 = 5**

**⇒ x² + x − 2 − 5 = 0**

**⇒ x² + x − 7 = 0**

**We have Δ = 1² − 4(1)(- 7) = 29 > 0.**

**There are two real solutions:**

**b. (x + 1)² = 3 − x²**

**⇒ x² + 2x + 1 = 3 − x²**

**⇒ x² + 2x + 1 + x² − 3 = 0**

**⇒ 2x² + 2x – 2 = 0**

**⇒ x² + x – 1 = 0**

**We have Δ = 1² − 4(1)(- 1) = 5 > 0**

**There are two real solutions:**

**Summary**

**Quadratic equation in standard form: ax² + bx + c = 0 (a ≠ 0)**

**How khổng lồ solve Quadratic Equations: (3 methods) Factoring, completing the square, and using the quadratic equation formula.**

**Remember: **the** discriminant** **Δ = b² − 4ac **is

**positive, there are two real solutions.****zero, there is one real solution.****negative, there is no real solution.****Now it’s your turn.Xem thêm: Nam Á Chủ Yếu Theo Tôn Giáo Nào, Dân Cư Dưới Đây**

I hope this post showed you how lớn solve quadratic equations using the 3 cool methods above.