Question

2. Solve the following quadratic equations using quadratic formula

Answer 1

3a²x² - abx - 2b² = 0
where, a = (3a²) , b = (-ab) , c = (-2b²)
D = b² - 4ac
D = (-ab)² - 4(3a²)(-2b²)
D = a²b² + 24a²b²
D = (5ab)²
√D = 5ab

X = [-b +- √D]/(2a]

X = (ab +- 5ab)/6a²

taking (-ve)
X = (ab - 5ab)/6a²
X = -4ab/6a² = -2b/3a

taking (+ve)
X = (ab + 5ab)/(6a²)
X = (6ab)/(6a²)
X = b/a

Answer 2

Compare given Quadratic equation

3a²x² - abx -2b² = 0 with

Ax² + Bx + C= 0 , we get

A = 3a² , B = -ab , C = -2b² ,

Discreminant ( D ) = B² - 4AC

= ( -ab )² - 4 × 3a² × ( -2b² )

= a²b² + 24a²b²

= 25a²b²

D = ( 5ab )²

By Quadratic Formula :

x = [ -B ± √D ]/2A

=> x = [ -(-ab) ± √( 5ab )² ]/( 2 × 3a² )

=> x = [ ab ± 5ab ]/6a²

Therefore ,

x = (ab+5ab)/6a² Or x = (ab-5ab)/6a²

x = 6ab/6a² Or x = -4ab/6a²

x = b/a Or x = -2b/3a

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