Quadratic Formula

Quadratic Formula helps to evaluate the solution of quadratic equations replacing the factorization method. A quadratic equation is of the form of ax2 + bx + c = 0, where a, b and c are real numbers, also called “numeric coefficients”Here x is an unknown variable, for which we need to find the solution.  

The Quadratic Formula to find the solution (roots) of the quadratic equation ax2 + bx + c = 0 is given by:

\(\LARGE x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\)

Here,

a, b, c = Constants (real numbers) 

a ≠ 0

x = Unknown, i.e. variable 

The sign of plus (+) and minus (-) represents that there are two solutions for quadratic equation, such as;

x1 = -b + √(b2-4ac)/2a and x2 = -b – √(b2-4ac)/2a

Both the solutions evaluated above are called the roots of the quadratic equation.

Also, read:

Example: Consider the quadratic equation x2-3x+2 = 0; substituting x=1 in LHS of the equation gives;

12-3+2 = 0 which is equal to RHS of the equation.

Similarly, substituting x = 2 in LHS of the equation also gives,

22-6+2 = 0 which is equal to RHS of the equation.

Here, 1 and 2 satisfy the equation x2-3x+2=0. Therefore, those are known as the solution of the quadratic equation or roots of the equation. This also means that numbers 1 and 2 are the zeros of the polynomial x2-3x+2. 

Roots of Quadratic Equation

We know that a second-degree polynomial will have at most two zeros. Therefore a quadratic equation will have at most two roots.

In general, if α is a root of the quadratic equation ax2+bx+c = 0, a ≠ 0; then, 2+bα+c = 0. We can also say that x = α is a solution of the quadratic equation or α satisfies the equation, ax2+bx+c=0.

Note:

  • Roots of the quadratic equation ax2+bx+c =0 are the same as zeros of the polynomial ax2+bx+c.
  • By splitting the middle term, we can factorise quadratic polynomials.

Example :  x2+5x+6  can be written as x2+2x+3x+6, x2+2x+3x+6 = x(x + 2) + 3(x + 2)= (x+2)(x + 3)

How to find the roots of quadratic equation?

Roots of the quadratic equation can be found by factorising the polynomial and equate it with zero.

Consider the quadratic equation 2x2-5x+2 = 0; which is of the form of ax2+ bx +c = 0

Here, -5x can be broken down into two parts like -4x and – x, as the multiplication of these parts result in 4xwhich is equal to a × c = 2x2 × 2 = 4x.Thus equation becomes:

2x2 – 5x + 2 = 2x2 – 4x – x + 2

= 2x(x – 2) -1 (x – 2)

= (2x – 1)(x – 2)

Therefore, 2x2-5x+2 = 0 is same as (2x – 1) (x – 2)=0

The values of x for which 2x2-5x+2 = 0 is the same as the values of x for which (2x – 1)(x – 2) = 0.

If (2x-1)(x – 2) = 0 ; then, either (2x – 1) = 0 or (x – 2) = 0

2x – 1 = 0 gives, 2x = 1 or  x = 1/2

x – 2 = 0 gives x = 2

Therefore, 1/2 and 2 are the roots of the equation 2x2-5x+2 = 0.

Quadratic Formula Derivation

Consider the equation ax2+bx+c = 0, a ≠ 0.

Dividing the equation by a gives,

 x2+ b/a x+c/a = 0

By using method of completing the square, we get

(x+b/2a)2 – (b/2a)2 + c/a = 0

(x+b/2a)2 – [(b2-4ac)/4a2]= 0

(x+b/2a)2 = (b2-4ac)/4a2

Roots of the equation are found by taking the square root of RHS. For that b2-4ac should be greater than or equal to zero.

When b2-4ac ≥ 0,

\( \left( x~+~\frac{b}{2a}\right) \) = ± \( \frac {\sqrt{b^2~-~4ac}}{2a}\)

\( x \) = \(\frac {-b~±~\sqrt{b^2~-~4ac}}{2a}\) —-(1)

Therefore roots of the equation are, \(\frac {-b~+~\sqrt{b^2~-~4ac}}{2a}\) and \(\frac {-b~-~\sqrt{b^2~-~4ac}}{2a}\)

The equation will not have real roots if b2-4ac < 0, because the square root is not defined for negative numbers in the real number system.

Equation (1) is a formula to find roots of the quadratic equation ax2+bx+c = 0, which is known as quadratic formula.

Also, the expression under the radicand, i.e. b2 – 4ac is called the determinant (D) of the given quadratic equation.

Based on the value of determinant, we can define the nature of roots.

If D = 0, the two roots are real and equal.

If D > 0, the roots are real and unequal.

If D < 0, the roots are not real, i.e. imaginary.

Examples

Example 1: Find the roots of the equation  x2-5x+6 = 0 using the quadratic formula.

Solution: Comparing the equation with ax2+bx+c = 0 gives,

a = 1, b = -5 and c = 6

b2-4ac = (-5)2-4×1×6 = 1

Roots of the equations are \(\frac {-b~+~\sqrt{b^2~-~4ac}}{2a}\)  = \( \frac {5~+~1}{2} \) = \( \frac {6}{2}\) = 3 and \(\frac {-b~-~\sqrt{b^2~-~4ac}}{2a}\)  = \( \frac {5~-~1}{2} \) = \( \frac {4}{2}\) = 2

Example 2: Find the roots of 4x2 + 3x + 5 = 0 using quadratic formula.

Solution:

Given quadratic equation is:

4x2 + 3x + 5 = 0

Comparing with the standard form ax2 + bx + c = 0,

a = 4, b = 3, c = 5

Determinant (D) = b2 – 4ac

= (3)2 – 4(4)(5)

= 9 – 80

= -71 < 0

That means, the roots are complex (not real).

Using quadratic formula,

x = [-b ± √(b2 – 4ac)]/ 2a

= [-3 ± √(-71)]/ 2(4)

= [-3 ± √(i2 71)]/ 8

= (-3 ± i√71)/8

Therefore, the complex roots of the given equation are x = (-3 + i√71)/8 and x (-3 – i√71)/8.

To know more about quadratic equations, download BYJU’S – The Learning App from Google play store.

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