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 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 = k
This principle can be extended to other perfect squares.
For example, if (x − 2)² = k then x − 2 = ±√k provided k > 0.
ExamplesExample 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. 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. To use the Null Factor law when solving equations, we must have one side of the equation equal lớn zero. 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. 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: #1 – Factoring the quadratic và then applying the Null Factor law
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.
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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
ExamplesExample 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.
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.
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 và is known as a repeated root.If Δ > 0, √Δ is a real number and so there are two distinct real roots
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.
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I hope this post showed you how lớn solve quadratic equations using the 3 cool methods above.
