Methods of Solving Quadratic Equations

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, \(\frac{1}{2}\)




9 Grade Math

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