Question

solve this quadratic formula or completion square


abx² - (a² + b²)x + ab = 0

Answer :- a/b , b/a



standard 10th

# content quality required!

fast guys ​

Answer 1

$\huge \bf \red{ \mid{ \overline{ \underline{ANSWER}}} \mid}$

→ abx²- (a²+b²)x+ab=0

» (by quadratic formula)

where, a'=ab,b'=-(a²+b²),c=ab

→ b'²-4ac=[-(a²+b²)]²-4(ab)(ab)

→ b'²-4ac=a⁴+b⁴-2(ab)1

→ b'²-4ac=[a⁴+b⁴-2(ab)²]

→ b'²-4ac=(a²-b²)1

» so, x= [-b ±√{b²-4ac}]/2a

→ x=[-{-(a²+b²)}±√{(a²-b²)/2ab

» taking (-ve)

→ x=[a²+b²-a²+b²)/2ab

→ x=2b²/2ab

→ x=b/a

→ taking (+ve)

→ x=[(a²+b²)+(a²-b²)]/2ab

→ x= (a²+b²+a²-b²)/2ab

→ x=2a²/2ab

» x=a/b

» so, x= a/b

and.

» x=b/a

Answer 2

$\huge \bf \red{ \mid{ \overline{ \underline{ANSWER}}} \mid}$

→ abx²- (a²+b²)x+ab=0

» (by quadratic formula)

where, a'=ab,b'=-(a²+b²),c=ab

→ b'²-4ac=[-(a²+b²)]²-4(ab)(ab)

→ b'²-4ac=a⁴+b⁴-2(ab)1

→ b'²-4ac=[a⁴+b⁴-2(ab)²]

→ b'²-4ac=(a²-b²)1

» so, x= [-b ±√{b²-4ac}]/2a

→ x=[-{-(a²+b²)}±√{(a²-b²)/2ab

» taking (-ve)

→ x=[a²+b²-a²+b²)/2ab

→ x=2b²/2ab

→ x=b/a

→ taking (+ve)

→ x=[(a²+b²)+(a²-b²)]/2ab

→ x= (a²+b²+a²-b²)/2ab

→ x=2a²/2ab

» x=a/b

» so, x= a/b

and.

» x=b/a