# CLASS-10SOLVING QUADRATIC EQUATION BY FORMULAE

Solving a Quadratic Equations by Formulae

Quadratic Formulae (Shreeedharacharya’s Rule) -

The roots of the equation ax² + bx + c = 0, where a ≠ 0, are given by –

- b ± √(b² - 4ac)

x = -------------------

2a

Proof Consider the equation ax² + bx + c = 0, where a, b, c є R, and a ≠ 0.

ax² + bx + c = 0

=> ax² + bx = - c

b               c

=> x² + ------ x = - -------     [Dividing throughout by a]

a               a

b               b               c           b

=> x² + ------- x + (-------)² = - ------ + (------)²

a              2a               a          2a

b

[Adding (------)² on both side]

2a

1         b           b              c          b

=> x² + 2 X ----- X ----- x + (-----)² = - ------ + (-----)²

2         a           2a             a          2a

b              b² - 4ac

=> (x + -------)² = (------------)

2a                4a²

b             √(b² - 4ac)

=> (x + -------) = ± --------------, when (b² - 4ac) ≥ 0

2a                 2a

-b          √(b² - 4ac)

=> x = ------- ± --------------

2a             2a

-b ± √(b² - 4ac)

=> x = -------------------, when (b² - 4ac) ≥ 0

2a

Thus the question ax² + bx + c = 0, a ≠ 0 has two roots given by –

-b ± √(b² - 4ac)

x = ------------------

2a

Example.1) Solve the following quadratic equation and calculate the answer correct to two decimal numbers :- x² - 5x – 10 = 0

Ans.) The given equation is x² - 5x – 10 = 0

This is of the form ax² + bx + c = 0, where a = 1, b = - 5, and c = -10

By, quadratic formulae, we have –

-b ± √(b² - 4ac)

x = ------------------

2a

- (-5) ± √{(-5)²- {4 X 1 X (-10)}

= ---------------------------------

2 X 1

5 ± √(25 + 40)         5 ± √65

= ----------------- = -----------

2                      2

5 + 8.06               5 – 8.06

=> x = ------------ or  x = ------------

2                        2

=> x = 13.06/2 or x = - 3.06/2

=> x = 6.53 or x = - 1.53

Solution set = {6.53, - 1.53}              (Ans.)

2x + 1         x + 3

Example.2) Solve the equation ---------- = ----------, write

x + 5          2x + 7

your answer correct to 2 decimal.

2x + 1           x + 3

Ans.) ----------- = -----------

x + 5            2x + 7

by cross multiplication, we have –

=> (2x + 1) (2x + 7) = (x + 3) (x + 5)

=> 4x² + 2x + 14x + 7 = x² + 3x + 5x + 15

=> 4x² + 16x + 7 = x² + 8x + 15

=> 3x² + 8x – 8 = 0

This is of the form ax²+ bx + c = 0, where a = 3, b = 8, and c = -8

By, quadratic formulae, we have –

-b ± √(b²- 4ac)

x = -------------------

2a

- 8 ± √[8² - {4 X 3 X (-8)}]

=> x = ------------------------------

2 X 3

- 8 ± √(64 + 96)

=> x = -------------------

6

- 8 ± √160            - 8 ± 4√10

=> x = --------------- = ----------------

6                       6

- 8 ± 4√10         - 8 ± (4 X 3.01)

=> x = -------------- = ------------------

6                       6

- 8 + 12.04               - 8 – 12.04

=> x = --------------- or x = ---------------

6                           6

4.04                - 20.04

=> x = ---------- or x = ------------

6                      6

=> x = 2.02/3 = 0.67 or x = - 3.34

Solution set = {0.67, - 3.34}          (Ans.)