We will discuss here about the methods of solving quadratic equations.
The quadratic equations of the form ax\(^{2}\) + bx + c = 0 is solved by any one of the following two methods (a) by factorization and (b) by formula.
(a) By factorization method:
In order to solve the quadratic equation ax\(^{2}\) + bx + c = 0, follow these steps:
Step I: Factorize ax\(^{2}\) + bx + c in linear factors by breaking the middle term or by completing square.
Step II: Equate each factor to zero to get two linear equations (using zero-product rule).
Step III: Solve the two linear equations. This gives two roots (solutions) of the quadratic equation.
Quadratic equation in general form is
ax\(^{2}\) + bx + c = 0, (where a ≠ 0) ………………… (i)
Multiplying both sides of, ( i) by 4a,
4a\(^{2}\)x\(^{2}\) + 4abx + 4ac = 0
⟹ (2ax)\(^{2}\) + 2 . 2ax . b + b\(^{2}\) + 4ac - b\(^{2}\) = 0
⟹ (2ax + b)\(^{2}\) = b\(^{2}\) - 4ac [on simplification and transposition]
Now taking square roots on both sides we get
2ax + b = \(\pm \sqrt{b^{2} - 4ac}\))
⟹ 2ax = -b \(\pm \sqrt{b^{2} - 4ac}\))
⟹ x = \(\frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\)
i.e., \(\frac{-b + \sqrt{b^{2} - 4ac}}{2a}\) or, \(\frac{-b - \sqrt{b^{2} - 4ac}}{2a}\)
Solving the quadratic equation (i), we have got two values of x.
That means, two roots are obtained for the equation, one is x = \(\frac{-b + \sqrt{b^{2} - 4ac}}{2a}\) and the other is x = \(\frac{-b - \sqrt{b^{2} - 4ac}}{2a}\)
Example to Solving quadratic equation applying factorization method:
Solve the quadratic equation 3x\(^{2}\) - x - 2 = 0 by factorization method.
Solution:
3x\(^{2}\) - x - 2 = 0
Breaking the middle term we get,
⟹ 3x\(^{2}\) - 3x + 2x - 2 = 0
⟹ 3x(x - 1) + 2(x - 1) = 0
⟹ (x - 1)(3x + 2) = 0
Now, using zero-product rule we get,
x - 1 = 0 or, 3x + 2 = 0
⟹ x = 1 or x = -\(\frac{2}{3}\)
Therefore, we get x = -\(\frac{2}{3}\), 1.
These are the two solutions of the equation.
(b) By using formula:
To form the Sreedhar Acharya’s formula and use it in solving quadratic equations
The solution of the quadratic equation ax^2 + bx + c = 0 are x = \(\frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\)
In words, x = \(\frac{-(coefficient of x) \pm \sqrt{(coefficient of x)^{2} – 4(coefficient of x^{2})(constant term)}}{2 × coefficient of x^{2}}\)
Proof:
Quadratic equation in general form is
ax\(^{2}\) + bx + c = 0, (where a ≠ 0) ………………… (i)
Dividing both sides by a, we get
⟹ x\(^{2}\) + \(\frac{b}{a}\)x + \(\frac{c}{a}\) = 0,
⟹ x\(^{2}\) + 2 \(\frac{b}{2a}\)x + (\(\frac{b}{2a}\))\(^{2}\) - (\(\frac{b}{2a}\))\(^{2}\) + \(\frac{c}{a}\) = 0
⟹ (x + \(\frac{b}{2a}\))\(^{2}\) - (\(\frac{b^{2}}{4a^{2}}\) - \(\frac{c}{a}\)) = 0
⟹ (x + \(\frac{b}{2a}\))\(^{2}\) - \(\frac{b^{2} - 4ac}{4a^{2}}\) = 0
⟹ (x + \(\frac{b}{2a}\))\(^{2}\) = \(\frac{b^{2} - 4ac}{4a^{2}}\)
⟹ x + \(\frac{b}{2a}\) = ± \(\sqrt{\frac{b^{2} - 4ac}{4a^{2}}}\)
⟹ x = -\(\frac{b}{2a}\) ± \(\frac{\sqrt{b^{2} - 4ac}}{2a}\)
⟹ x = \(\frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\)
This is the general formula for finding two roots of any quadratic equation. This formula is known as quadratic formula or Sreedhar Acharya’s formula.
Example to Solving quadratic equation applying Sreedhar Achary’s formula:
Solve the quadratic equation 6x\(^{2}\) - 7x + 2 = 0 by applying quadratic formula.
Solution:
6x\(^{2}\) - 7x + 2 = 0
First we need to compare the given equation 6x\(^{2}\) - 7x + 2 = 0 with the general form of the quadratic equation ax\(^{2}\) + bx + c = 0, (where a ≠ 0) we get,
a = 6, b = -7 and c =2
Now apply Sreedhar Achary’s formula:
x = \(\frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\)
⟹ x = \(\frac{-(-7) \pm \sqrt{(-7)^{2} - 4 ∙ 6 ∙ 2}}{2 × 6}\)
⟹ x = \(\frac{7 \pm \sqrt{49 - 48}}{12}\)
⟹ x = \(\frac{7 \pm 1}{12}\)
Thus, x = \(\frac{7 + 1}{12}\) or, \(\frac{7 - 1}{12}\)
⟹ x = \(\frac{8}{12}\) or, \(\frac{6}{12}\)
⟹ x = \(\frac{2}{3}\) or, \(\frac{1}{2}\)
Therefore, the solutions are x = \(\frac{2}{3}\) or, \(\frac{1}{2}\)
Quadratic Equation
Introduction to Quadratic Equation
Formation of Quadratic Equation in One Variable
General Properties of Quadratic Equation
Methods of Solving Quadratic Equations
Examine the Roots of a Quadratic Equation
Problems on Quadratic Equations
Quadratic Equations by Factoring
Word Problems Using Quadratic Formula
Examples on Quadratic Equations
Word Problems on Quadratic Equations by Factoring
Worksheet on Formation of Quadratic Equation in One Variable
Worksheet on Quadratic Formula
Worksheet on Nature of the Roots of a Quadratic Equation
Worksheet on Word Problems on Quadratic Equations by Factoring
From Methods of Solving Quadratic Equations to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
Mar 02, 24 05:31 PM
Mar 02, 24 04:36 PM
Mar 02, 24 03:32 PM
Mar 01, 24 01:42 PM
Feb 29, 24 05:12 PM
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.