Quadratic Equation
Introduction
A quadratic equation is an equation of the form ax2 + bx + c = 0 where a, b, and c are constants and a ≠ 0.
The origin of word "quadratic" is Latin. In Latin, "quadratic" is used for "square". Since the highest power of the unknown variable that appears in the equation is square(power 2), hence similar equations came to be known as quadratic equation.
The standard form of a quadratic equation is ax2 + bx + c = 0, where
◾ a, b and c are constants and
◾ a is not equal to 0 (zero).
Here are some of the examples of standard form of a quadratic equation:
| Equation | Coefficient |
|---|---|
| x2 + 2x + 1 = 0 | where a = 1, b = 2 and c = 1 |
| x2 + 5x + 11 = 0 | where a = 1, b = 5 and c = 11 |
◾ When b is not equal to zero
| Some of the examples of complete form of a quadratic equation |
|---|
| ◾ x2 + 2x + 1 = 0 where a = 1, b = 2 and c = 1 |
| ◾ x2 + 5x + 11 = 0 where a = 1, b = 5 and c = 11 |
◾ When b is equal to zero, the equation is known as pure or incomplete quadratic equation in x.
| Some of the examples of Pure or Incomplete form of a quadratic equation |
|---|
| ◾ 6x2 – 24 = 0 where b = 0 |
| ◾ 6x2 – 11 = 0 where b = 0 |
The Roots or solution of the quadratic equation of ax2 + bx + c =0 are the values of the variable 'x' which satisfy the quadratic equation i.e. make the ax2+bx+c equals to zero.
For example:
x2 + 5x - 50 = 0
x2 - 5x + 10x -50 = 0
x(x - 5) + 10(x - 5) = 0
(x - 5) (x + 10) = 0
x = 5 and x = -10
As you can see that putting either 5 or -10 in place of x makes the quadratic equation x2 + 5x - 10 equals to zero. Hence, 5 and -10 are the roots of the quadratic equation x2 + 5x - 50 = 0.
There are three methods for solving a quadratic equation:
◾ By doing factorization
◾ By completing the square
◾ By using the quadratic formula
Step by step process to solve the quadratic equation by factorization method:
Step 1: Rearrange the equation in standard form which is ax2 + bx + c = 0. If right hand side is not zero, take it to left hand side and make the right hand side zero.
Step 2: Fully factorise the left hand side.
Step 3: Use the Null Factor law: If pq = 0 then p = 0 or q = 0.
Step 4: Solve the resulting linear equations by equating each of the linear factor to zero. These values of x will be the solution of the quadratic equation.
Example: Solve 3x2 + 5x - 5 = -3
Step 1: Rearranging in standard form
3x2 + 5x -5 + 3 = 0
3x2 + 5x - 2 = 0
Step 2: Factorize the left hand side
3x2 + 6x – x – 2 = 0
3x(x + 2) -1(x + 2) = 0
(3x – 1) (x + 2) = 0
Step3: Equate each of the linear factor to zero.
3x – 1 = 0 or x + 2 = 0
3x = 1 or x = -2
x = 1/3, x = - 2 are the roots of the Equation.
Pitfall:
You may be tempted to divide both sides by an expression involving x. If you do this then you only get one solution of equation (one value or one root of x) and may lose the other solution (value of x).
For example: consider x2 = 7x
Correct solution:
x2 = 7x
x2-7x=0
x(x-7)=0
x=0 and X=7
Incorrect Solution:
x2 = 7x
Dividing both sides by x, we get
X=7
above is incorrect way of solving the equation, since we could not find the other value of x which is zero.
As you would be aware by now that all quadratic equations cannot be factorised easily. For example, x2 + 4x + 1 cannot be factorised by simple factorisation. It means we cannot write x2 + 4x + 1 in the form (x - a)(x - b) where a, b are rational numbers.
There is an alternate way to solve equations like x2 + 4x + 1 = 0 that is by completing the square.
Equations of the form ax2 + bx + c = 0 can be converted to the form (x + p)2 = q . It is easy to find the solutions in that way.
Step by step process to solve the quadratic equation by completing the square:
Step I: Arrange the quadratic equation in standard form of ax2 + bx + c = 0.
Step II: now, divide both sides of the equation by the co-efficient of x2 if it is not already 1.
Step III: Shift the constant term to the right hand side.
Step IV: Add the square of one-half of the co-efficient of x to L.H.S. and R.H.S.
Step V: Write the L.H.S as complete square and simplify the R.H.S.
Step VI: Solve for x by taking the square root of L.H.S. and R.H.S.
Let us solve quadratic equation -3x2 + 12x + 5 = 0 by "completing the square"
-3x2 + 12x + 5 = 0
x2 - 4x - (5/3) = 0
x2 - 4x = (5/3)
x2 - 4x + 22 = (5/3) + 22
(x - 2)2 = (17/3)
x - 2 = ± √(17/3)
x = 2 ± √(17/3)
x1 = 2 + √(17/3)
x2 = 2 - √(17/3)
So, Here x1 and x2 are the roots of equation.The quadratic formula, which also may be used to solve any quadratic equation, results from solving the quadratic equation ax2 + bx + c = 0 , a ≠ 0 for x by completing the square.
There are certain cases when solving a quadratic equation by factorisation or completing the square is time consuming, lengthy or difficult. In such cases, we use the quadratic formula to solve the quadratic equation.
Step by step process to solve the quadratic equation by quadratic formula:
Step I: Arrange the Quadratic Equation in the standard form of ax2 + b x + c = 0.
Step II: Compare the quadratic equation which is to be solved with the standard quadratic equation and find out the values of coefficients a, b, and c.
Step III: Put these values of a, b, and c in Quadratic formula.
| Roots(x1, x2) = | −b ± √ b2 − 4ac |
| 2a |
As you can notice in the formula. Quadratic formula calculates two values of x: x1 and x2, where
| x1 = | −b + √ b2 − 4ac |
| 2a |
| x2 = | −b - √ b2 − 4ac |
| 2a |
These two values of x for which ax2 + bx + c = 0 holds true, are called solutions of Quadratic Equation, also called roots of Quadratic Equation.
Step I: - Convert the quadratic equation you wish to solve in Standard Form of Quadratic Equation, ax2 + bx + c = 0
For example, if you have a quadratic equation in the form of x2 - 10x = -24, then convert it into Standard Form of Quadratic Equation.
x2 -10x = -24 is converted into x2 - 10x + 24 = 0
Step II: - Find value of coefficient a, b and c by comparing it with Standard Form of Quadratic Equation ax2 + bx + c = 0
For example, comparing x2 - 10x + 24 = 0 with ax2 + bx + c = 0, we get
a = 1,
b = -10,
c = 24
Step III:
| Roots(x1, x2) = | −b ± √ b2 − 4ac |
| 2a |
| x1 = | −b + √ b2 − 4ac |
| 2a |
| = | −(-10) + √ (-10)2 − 4(1)(24) | = 6 |
| 2(1) |
| x2 = | −b - √ b2 − 4ac |
| 2a |
| = | −(-10) - √ (-10)2 − 4(1)(24) | = 4 |
| 2(1) |
Discriminant
We have learnt that the quadratic formula is
| Roots(x1, x2) = | −b ± √ b2 − 4ac |
| 2a |
In the quadratic formula above, the quantity "b2 - 4ac" which is under the square root sign is called the discriminant of the quadratic equation.
Discriminant = b2 - 4ac
The expression "b2 – 4ac" tells about the nature of the roots of the quadratic equation. The roots may be real, equal or imaginary.
There are three possible cases:
◾ If b2 – 4ac < 0, then roots will be imaginary and unequal.
◾ If b2 – 4ac = 0, then roots will be real, equal and rational . (This means the left hand side of the equation is a perfect square).
◾ If b2 – 4ac > 0, then the roots are real and unequal.
If b2 – 4ac > 0, then the roots are real and unequal and their are two possibilities - Here the roots could be rational or irrational
◾ b2 – 4ac is a perfect square, the roots are real, rational and unequal . (This mean the equation can be solved by the factorization).
◾ b2 – 4ac is a not perfect then roots are real, irrational and unequal.
Summary
| Discriminant value Cases | Roots of quadratic | Factorisation of quadratic |
|---|---|---|
| Discriminant value > 0 | two real distinct roots | two distinct linear factors |
| Discriminant value = 0 | two identical real roots | two identical linear factors |
| Discriminant value < 0 | No real roots | Unable to factorise |
Case I- When Discriminant > 0
For a quadratic equation x2 + 7x + 4 = 0, determine the nature of roots by its determinant
Answer
x2 + 7x + 4 = 0
Discriminant = b2 – 4ac
Here, a = 1, b = 7, and c = 4
Discriminant = 72 – 4 x 1 x 4 = 49 - 16 = 33
Discriminant > 0, Therefore there are two real roots
Case II- When Discriminant = 0
For a quadratic equation x2 + 6x + 9 = 0, determine the nature of roots by its determinant
Answer
x2 + 6x + 9 = 0
Discriminant = b2 – 4ac
Here, a = 1, b = 6, and c = 9
Discriminant = 62 – 4 x 1 x 9 = 36 - 36 = 0
Discriminant = 0, Therefore there are two identical real roots
Case III- When Discriminant < 0
For a quadratic equation x2 + 4x + 4 = 0, determine the nature of roots by its determinant
Answer
x2 + 4x + 5 = 0
Discriminant = b2 – 4ac
Here, a = 1, b = 4, and c = 5
Discriminant = 42 – 4 x 1 x 5 = 16 - 20 = -4
Discriminant < 0, Therefore there are no real roots