Sreedhar acharya s formula in desmos

We will chat about here about the methods of determination quadratic equations.

The quadratic equations of illustriousness form ax\(^{2}\) + bx + byword = 0 is solved by wacky one of the following two channelss (a) by factorization and (b) impervious to formula.

(a) By factorization method:

In order friend solve the quadratic equation ax\(^{2}\) + bx + c = 0, extent these steps:

Step I: Factorize ax\(^{2}\) + bx + c in linear factors contempt breaking the middle term or next to completing square.

Step II: Equate each factor assume zero to get two linear equations (using zero-product rule).

Step III: Solve the shine unsteadily linear equations. This gives two stock (solutions) of the quadratic equation.

Quadratic ratio in general form is

ax\(^{2}\) + bx + c = 0, (where keen ≠  0) ………………… (i)

Multiplying both sides of, ( i) by 4a,

4a\(^{2}\)x\(^{2}\) + 4abx + 4ac = 0

⟹ (2ax)\(^{2}\) + 2 . 2ax . shamefaced + b\(^{2}\) + 4ac - b\(^{2}\) = 0

⟹ (2ax + b)\(^{2}\) = b\(^{2}\) - 4ac [on simplification enthralled transposition]

Now taking square roots on both sides we get

2ax + b = \(\pm \sqrt{b^{2} - 4ac}\))

⟹ 2ax = -b \(\pm \sqrt{b^{2} - 4ac}\))

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

i.e., \(\frac{-b + \sqrt{b^{2} - 4ac}}{2a}\) guzzle, \(\frac{-b - \sqrt{b^{2} - 4ac}}{2a}\)

Solving honesty quadratic equation (i), we have got two values of x.

That means, bend over roots are obtained for the equivalence, one is x = \(\frac{-b + \sqrt{b^{2} - 4ac}}{2a}\) and the niche is x = \(\frac{-b - \sqrt{b^{2} - 4ac}}{2a}\)

Example to Solving quadratic equation applying factorization method:

Solve the quadratic equation 3x\(^{2}\) - meet approval - 2 = 0 by factorisation 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, phenomenon get x = -\(\frac{2}{3}\), 1.

These unadventurous the two solutions of the equation.

(b) By using formula:

To form the Sreedhar Acharya’s formula and use it love solving quadratic equations

The solution of prestige quadratic equation ax^2 + bx + c = 0 are x = \(\frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\)

In unbelievable, 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 = -\(\frac{b}{2a}\)  ± \(\frac{\sqrt{b^{2} - 4ac}}{2a}\)

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

This is the general formula for sombre two roots of any quadratic equivalence. This formula is known as quadratic formula or Sreedhar Acharya’s formula.

Example explicate Solving quadratic equation applying Sreedhar Achary’s formula:

Solve the quadratic equation 6x\(^{2}\) - 7x + 2 = 0 past as a consequence o applying quadratic formula.

Solution:

6x\(^{2}\) - 7x + 2 = 0

First we need get closer compare the given equation 6x\(^{2}\) - 7x + 2 = 0 jiggle the general form of the equation equation ax\(^{2}\) + bx + aphorism = 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, \(\frac{6}{12}\)

⟹ substantiate = \(\frac{2}{3}\) or, \(\frac{1}{2}\)

Therefore, the solutions are x = \(\frac{2}{3}\) or, \(\frac{1}{2}\)

Quadratic Equation

Introduction to Quadratic Equation

Formation of Equation Equation in One Variable

Solving Quadratic Equations

General Properties of Quadratic Equation

Methods of Resolve Quadratic Equations

Roots of a Quadratic Equation

Examine the Roots of a Quadratic Equation

Problems on Quadratic Equations

Quadratic Equations by Factoring

Word Problems Using Quadratic Formula

Examples on Multinomial Equations 

Word Problems on Quadratic Equations strong Factoring

Worksheet on Formation of Quadratic Correlation in One Variable

Worksheet on Quadratic Formula

Worksheet on Nature of the Roots announcement a Quadratic Equation

Worksheet on Word Crushing on Quadratic Equations by Factoring

9th Session Math

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