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Today you are going to see 3 methods to solve quadratic equations that you need to lớn know.
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And these methods actually work.
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.
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 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. 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. 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 & 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 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
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 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.
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
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.
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I hope this post showed you how to lớn solve quadratic equations using the 3 cool methods above.
