Proof of the Quadratic Formulas and Questions
The an analytical proof of the quadratic formulas used to solve quadratic equations is presented. Examples on how to use the quadratic formulas and the discriminant to solve various questions related to quadratic equation are also presented with detailed explanations.
Analytical Proof of the Quadratic Formulas
A quadratic equation in the standard form is given bya x^{2} + b x + c = 0
where a, b and c are constants with a not equal to zero.
Solve the above equation to find the quadratic formulas

Given
a x^{2} + b x + c = 0

Divide all terms by a
x^{2} + b / a x + c / a = 0

Subtract c / a from both sides
x^{2} + (b / a) x + c / a  c/ a = 0  c / a

and simplify
x^{2} + (b / a) x =  c / a

Add (b / 2a)^{2} to both sides
x^{2} + (b / a) x + (b / 2a)^{2} =  c / a + (b / 2a)^{2}

and complete the square on the left hand side of the equation
(x + b / 2a )^{2} =  c / a + (b / 2a)^{2}

Group the two terms on the right side of the equation
(x + b / 2a)^{2} = (b^{2}  4 a c) / (4 a^{2} )

Solve by taking the square root
x + b / 2a = ~+mn~ √( (b^{2}  4 a c ) / (4 a^{2}) )

Solve for x to obtain two solutions
x =  b / 2a ~+mn~ √( (b^{2}  4 a c ) / (4 a^{2}) )

The term √( (b^{2}  4 a c ) / (4 a^{2}) ) may be simplified as follows
√( (b^{2}  4 a c ) / (4 a^{2}) ) = √(b^{2}  4 a c) / (2 a)

Since 2  a  = 2 a for a > 0 and 2  a  =  2 a for a < 0, the two solutions to the quadratic equation may be written
x_{1} = (  b + √( b^{2}  4 a c)) / (2 a)
x_{2} = (  b  √ ( b^{2}  4 a c)) / (2 a)
Δ = b^{2}  4 a c is called the discriminant 
The term b^{2}  4 a c which is under the square root in both solutions is called the discriminant of the quadratic equation. It can be used to determine the number and nature of the solutions of the quadratic equation. 3 cases are possible
case 1: If Δ > 0 , the equation has 2 solutions.
case 2: If Δ = 0 , the equation has one solutions of multiplicity 2.
case 3: If Δ < 0 , the equation has 2 complex solutions.
Questions on Quadratic Equations
Question 1
Solve the equationSolution to Question 1
We first identify the coefficients a, b and c.a = 2 , b =  7 and c = 4
The discriminant Δ = b^{2}  4 a c is now calculated.
Δ = b ^{2}  4 a c = (7) ^{2}  4 (2)(4) = 17
The discriminant is positive and therefore the equation has two real solutionsUse given by the quadratic formulas:
x _{1} = ( b + √( b ^{2}  4 a c)) / (2 a) = (7 + √(17)) / 4
= 7 / 4 + √(17) / 4
x _{2} = ( b  √ ( b ^{2}  4 a c)) / (2 a) = (7  √(17)) / 4
= 7 / 4  √(17) / 4
Question 2
Solve the equationSolution to Question 2
We first identify the coefficients a, b and c.a = 1 , b =  2 and c = 5
The discriminant Δ = b^{2}  4 a c is now calculated.
Δ = b ^{2}  4 a c = (2) ^{2}  4 (1)(5) =  16
The discriminant is negative and therefore the equation has two complex solutions. In complex numbers, we define the imaginary unit i as
i = √(1)
The Solutions to the equation are given by
x _{1} = ( b + √( b ^{2}  4 a c)) / (2 a) = (2 + √(16)) / 2
x _{2} = ( b  √ ( b ^{2}  4 a c)) / (2 a) = (2  √(16)) / 2
We now simplify √(16) as follows
√(16) = √(1)√(16) = 4 i , where i is the imaginary unit defined above.
We now simplify the two solutions
x _{1} = (2 + √(16)) / 2 = (2 + 4 i) / 2
= 1 + 2 i
x _{2} = (2  √(16)) / 2 = (2  4 i) / 2
= 1  2 i
Question 3
How many solutions does the equationSolution to Question 3
We first write the given equation in standard form (with the zero on the right) in the form a x^{2} + b x + c = 0.2x ^{ 2} + 3x + 9 = 0
We now identify the coefficients a, b and c.
a = 2 , b = 3 and c = 9
The sign of the discriminant Δ = b^{2}  4 a c gives the number of solutions to the equation.
Δ = b ^{2}  4 a c = (3) ^{2}  4 (2)(9) =  63
The discriminant Δ is negative and therefore the equation has two complex solutions.
Question 4
Find all values for the parameter m so that the equationSolution to Question 4
We first identify the coefficients a, b and c.a = 1 , b = m and c = 4
The sign of the discriminant Δ = b^{2}  4 a c gives the number of solutions to the equation.
Δ = b ^{2}  4 a c = m ^{2}  4 (1)(4) = m ^{2}  16
For the equation to have two real solutions, the discriminant Δ must be positive. Hence we need to solve the following inequality in m
m ^{2}  16 > 0
Factor the right side of the inequality
(m  4)(m + 4) > 0
The values of m for which the given equation has two real solutions are the solutions to the above inequality and are given by the interval
(  ∞ ,  4) ∪ (4 , ∞)
Question 5
Find all values for the parameter k so that the equationSolution to Question 5
We first write the equation in standard form. x ^{ 2} + (k + 1) x  2  k = 0
We now identify the coefficients a, b and c.
a =  1 , b = k + 1 and c =  2  k
The discriminant Δ = b^{2}  4 a c is given by
Δ = b ^{2}  4 a c = (k + 1) ^{2}  4 (  1)(  2  k) = (k + 1) ^{2}  8  4 k
For the equation to have one solution, the discriminant Δ must be equal to zero. Hence we need to solve the following equation in k
(k + 1) ^{2}  8  4 k = 0
Expand the above equation, simplify and rewrite it as
k ^{2}  2 k  7 = 0
Solve the above quadratic equation in k using the quadratic formulas to obtain the two values of k for which the given equation has one solution only.
Δ = (2) ^{2}  4 (1)(7) = 32
k _{1} = ( 2 + √32) / (2 * 1)
k _{2} = ( 2  √32) / (2 * 1)
which simplifies to
k _{1} = 1 + 2√2
and k _{2} = 1  2√2
NOTE: As an exercise, use one value of k found above in the given equation and solve it; you must obtain one solution only.
More References and links
Solve Equations, Systems of Equations and Inequalities .Step by Step Quadratic Equation Solver.