1. If Δ > 0, the equation has 2 real solutions. (see example 1 below)
   
2. If Δ = 0, the equation has 1 real solution. (see example 2 below)
   
3. If Δ < 0, the equation has 2 conjugate complex solutions. (see example 3 below)

Questions with Solutions

Question 1
Use the quadratic formulas and the discriminant to find all solutions to the quadratic equation given below.
x ² + 3 x = 4
Solution to Question 1

• Given
  x ² + 3 x = 4
  
• Rewrite the given equation with its right term equal to zero.
  x ² + 3 x - 4 = 0
  
• Find the discriminant Δ= b² - 4 a c
  Δ= b² - 4 a c = 3² - 4(1)(-4) = 25
  
• Since the discriminant is positive, the equation has two real solutions given by.
  x₁ = ( -3 + √25 ) / (2*1) = ( -3 + 5 ) / 2 = 1
  x₂ = ( -3 - √25) / (2*1) = (-3 - 5) / 2 = - 4

1. x = 1
   Left side of the equation = x ² + 3 x = 1 ² + 3(1) = 1 + 3 = 4
   Right side of the equation = 4.
   
2. x = -4
   Left side of the equation = (-4) ² + 3(-4) = 16 - 12 = 4
   Right side of the equation = 4.

Question 2
Use the quadratic formulas and the discriminant to find all solutions to the quadratic equation given below.
x ² / 3 + 3 = 2 x
Solution to Question 2

• Given
  x ² / 3 + 3 = 2 x
  
• Eliminate the denominator by multiplying all terms in the equation by 3.
  3 (x ² / 3 + 3) = 3 × 2x
  
• Simplify and rewrite the equation with the right term equal to zero.
  x ² - 6 x + 9 = 0
  
• Use the quadratic formula. The discriminant Δis given by
  Δ= b² - 4 a c
  = (-6)² - 4(1)(9) = 0
  
• Since the discriminant is equal to zero, the two formulas giving the two solutions of the quadratic equation become one solution given by
  x = - b / (2 a) = -(-6) / (2*1) = 3
  

1. x = 3
   Left side of the equation = x ² / 3 + 3 = 3 ² / 3 + 3 = 6
   Right side of the equation = 2 x = 2(3) = 6.

Question 3
Use the quadratic formulas and the discriminant to find all solutions to the quadratic equation given below.
x ² - 4 x + 13 = 0 Solution to Question 3

• Given
  x ² - 4 x + 13 = 0
  
• The discriminant Δis given by
  Δ= b² - 4 a c
  = (-4)² - 4(1)(13) = -36
  
  
• Since the discriminant is negative, the square root of the discriminant is a pure imaginary number.
  sqrt(D) = √(-36) = √(-1)√(36) = 6i
  where i is the imaginary unit defined as i = √(-1).
  
• Use the quadratic formulas to find the two solutions.
  x₁ = (4 + 6 i)) / (2*1) = 2 + 3 i
  x₂ = (4 - 6 i) / (2*1) = 2 - 3 i

1. x = 2 + 3 i
   Left side of the equation = x ² - 4 x + 13 = (2 + 3 i) ² - 4(2 + 3 i) + 13
   = 4 - 9 + 12 i - 8 - 12 i + 13 = 0
   Right side of the equation = 0
   
2. x = 2 - 3 i
   Left side of the equation = x ² - 4 x + 13 = (2 - 3 i) ² - 4(2 - 3 i) + 13
   = 4 - 9 - 12 i - 8 + 12 i + 13 = 0
   Right side of the equation = 0

Question 4
Find all values of the parameter m in the quadratic equation
x ² + m x + 1 = 0
such that the equation has

1. one solution,
2. 2 real solutions,
3. 2 complex solutions.

• a) Given
  x ² + m x + 1 = 0
  
• Find the discriminant Δ = b² - 4ac
  Δ = b² - 4ac = m² - 4(1)(1) = m² - 4
  
• For the equation to have one solution, the discriminant has to be equal to zero.
  m² - 4 = 0
  
• The equation m² - 4 = 0 has two solutions.
  m = 2
  m = -2
  Below is the graph of the expression in the left side of the given equation for m = 2 and m = -2.
  Note that in each case, the graph has 1 x intercept only, hence one real solution to the equation.
  

  

• b) For the equation to have 2 real solution, the discriminant has to be greater than zero.
  m² - 4 > 0
  
• The inequality m² - 4 > 0 has the following solution set.
  (-∞ , -2) U (2 , +∞) &infty;
  Below is the graph of the expression in the left side of the given equation for m = 5 and m = -3.
  Note that in each case, the graph has 2 x-intercepts, hence the given equation has 2 real solutions.
  

  

• c) For the equation to have 2 complex solution, the discriminant has to be less than zero.
  m² - 4 < 0
  
• The inequality m² - 4 < 0 has the following solution set.
  (-2 , 2)

More Questions with Detailed Solutions