Quadratic Formula helps to evaluate the solution of quadratic equations replacing the factorization method. A quadratic equation is of the form of ax^{2} + 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 ax^{2} + 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;
x_{1} = -b + √(b^{2}-4ac)/2a and x_{2} = -b – √(b^{2}-4ac)/2a
Both the solutions evaluated above are called the roots of the quadratic equation.
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Example: Consider the quadratic equation x^{2}-3x+2 = 0; substituting x=1 in LHS of the equation gives;
1^{2}-3+2 = 0 which is equal to RHS of the equation.
Similarly, substituting x = 2 in LHS of the equation also gives,
2^{2}-6+2 = 0 which is equal to RHS of the equation.
Here, 1 and 2 satisfy the equation x^{2}-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 x^{2}-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 ax^{2}+bx+c = 0, a ≠ 0; then, aα^{2}+bα+c = 0. We can also say that x = α is a solution of the quadratic equation or α satisfies the equation, ax^{2}+bx+c=0.
Note:
- Roots of the quadratic equation ax^{2}+bx+c =0 are the same as zeros of the polynomial ax^{2}+bx+c.
- By splitting the middle term, we can factorise quadratic polynomials.
Example : x^{2}+5x+6 can be written as x^{2}+2x+3x+6, x^{2}+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 2x^{2}-5x+2 = 0; which is of the form of ax^{2}+ bx +c = 0
Here, -5x can be broken down into two parts like -4x and – x, as the multiplication of these parts result in 4x^{2 }which is equal to a × c = 2x^{2} × 2 = 4x^{2 }.Thus equation becomes:
2x^{2} – 5x + 2 = 2x^{2} – 4x – x + 2
= 2x(x – 2) -1 (x – 2)
= (2x – 1)(x – 2)
Therefore, 2x^{2}-5x+2 = 0 is same as (2x – 1) (x – 2)=0
The values of x for which 2x^{2}-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 2x^{2}-5x+2 = 0.
Quadratic Formula Derivation
Consider the equation ax^{2}+bx+c = 0, a ≠ 0.
Dividing the equation by a gives,
x^{2}+ 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} – [(b^{2}-4ac)/4a^{2}]= 0
(x+b/2a)^{2} = (b^{2}-4ac)/4a^{2}
Roots of the equation are found by taking the square root of RHS. For that b^{2}-4ac should be greater than or equal to zero.
When b^{2}-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 b^{2}-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 ax^{2}+bx+c = 0, which is known as quadratic formula.
Also, the expression under the radicand, i.e. b^{2} – 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 x^{2}-5x+6 = 0 using the quadratic formula.
Solution: Comparing the equation with ax^{2}+bx+c = 0 gives,
a = 1, b = -5 and c = 6
b^{2}-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 4x^{2} + 3x + 5 = 0 using quadratic formula.
Solution:
Given quadratic equation is:
4x^{2} + 3x + 5 = 0
Comparing with the standard form ax^{2} + bx + c = 0,
a = 4, b = 3, c = 5
Determinant (D) = b^{2} – 4ac
= (3)^{2} – 4(4)(5)
= 9 – 80
= -71 < 0
That means, the roots are complex (not real).
Using quadratic formula,
x = [-b ± √(b^{2} – 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.
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