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.)