How to solve quadratic equations – 3 methods – Step by step

quadratic equations

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

And these methods actually work. 

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

Quadratic equations definition

quadratic equations

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 and error?

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

How to solve quadratic equations

Quadratic equations of the form 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.

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 < 0.


Example 2: Solve for x: 

a) (x − 2)² = 25    (we do 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 and then apply the Null Factor law.

#1 –  Factoring the quadratic and 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 to zero.

  1. If necessary, rearrange the equation so one side is zero.
  2. Fully factorize the other side (usually the LHS)
  3. Use the Null Factor law: if a × b = 0 then a = 0 or b = 0.
  4. Solve the resulting linear equations.
  5. 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) = − 8
  • find the factors of ac which add to b: b = 2 so, these factors are 4 and − 2 (sum = 2 and product = − 8)
  • replace 2x by 4x − 2x
  • complete 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 and − 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.

Check: If x = − 2 then LHS = 2 and 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 do we add on to make a perfect square?

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

Consider x² + 6x + 2 = 0.

The first step is to keep the terms containing x on the LHS and 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 add 3² or 9 to 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 to 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 add (-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, and completing the square is rather tedious.

Consequently, the quadratic formula has been developed.

quadratic equation formula

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

But it’s not always worked out like 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, and 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 and is known as a repeated root.
  • If Δ > 0, √Δ is a real number and so there are two distinct real roots
  • If Δ < 0, √Δ does not exist and so we have no real solutions. We get a pair of complex solutions.
  • If a, b and 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 < 0.

Therefore, there are no real roots.


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 to solve:

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


Example 5: Use the quadratic equation formula to 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 to solve Quadratic Equations: (3 methods) Factoring, completing the square, and using the quadratic equation formula.

Remember: the discriminant Δ = b² − 4ac is

  1. positive, there are two real solutions.
  2. zero, there is one real solution.
  3. negative, there is no real solution. 

Now it’s your turn.

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

Now I’d like to hear from you:

Which method from today’s post do you like most?

Find more about Complex Solutions of Quadratic equations.

Find more about Mathematics

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