Giải X2

     

Today you are going to see 3 methods to solve quadratic equations that you need to lớn know.

Bạn đang xem: Giải x2

And these methods actually work. 

*
*
First of all, you need to know what quadratic equations are.


Quadratic equations definitionHow lớn solve quadratic equationsQuadratic equations of the size 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 and 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 do we find these solutions without using trial & error?

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

How to solve quadratic equations

Quadratic equations of the size 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 = k

*
*

This principle can be extended khổng lồ 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 vày 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 & then applying the Null Factor law

The Null Factor law states that:

When the product 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 add to b: b = 2 so, these factors are 4 và − 2 (sum = 2 & product = − 8)replace 2x by 4x − 2xcomplete the factorization

In 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: Hướng Dẫn Cách Ghép Video Trên Điện Thoại, Please Wait

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 vày we địa chỉ cửa hàng on lớn make a perfect square?

Halve the coefficent of x. Add the square of this number lớn both sides of the equation.

Consider x² + 6x + 2 = 0.

The first step is to lớn keep the terms containing x on the LHS & 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 lớn both sides of the equation.

So, x² + 6x + 9 = − 2 + 9 

Therefore, (x + 3)² = 7

⇒ x + 3 = ±√7

⇒ x = − 3 ±√7 

Examples

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 lớn 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 khổng lồ the RHS: x² − 2x = − 7

We showroom (-2/2)² to 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, và 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 khổng lồ 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, và do the calculations.

Therefore,

*
*

b) 2x² + 3x − 4 = 0

We have a = 2, b = 3 and c = −4. We plug in the values of a, b & c, and 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, và this further justifies the fact that x² + 2x + 5 = 0 has no real solutions.

The discriminant, Δ (delta)

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:

If Δ = 0, x = −b/(2a) is the only solution & 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 and 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 và plug in the values of a, b, and c, then vày the calculations.

*
*

Example 5: Use the quadratic equation formula to lớn 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 lớn solve Quadratic Equations: (3 methods) Factoring, completing the square, và 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: Nêu 5 Ví Dụ Về Hiện Tượng Vật Lý Và Hiện Tượng Hóa Học, Hiện Tượng Vật Lý Là Gì

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