for which condition, the quadratic equation (a² + b²) x² + 2 (b c – a d) x + (c² + d²) = 0 has real and equal roots?
...... plz ans...I have test today...so plz give correct ans...plz...
Answers
Answer:
.e. D = 0
= > b² - 4 ac = 0
From question we have given :
b = 2 ( b c - a d )
a = a² + b²
c = c² - d²
Now put all value in condition.
= > ( 2 ( b c - a d )² - 4 * ( a² + b² ) ( c² + d² ) = 0
= > 4 b² c² +4 a² d² - 8 b c a d - 4 * ( a² c² + a² d²+ b² c² + b² d² ) = 0
= >4 b² c² +4 a² d² - 8 b c a d - 4 a² c² - 4 a² d² - 4 b² c² - b² d² = 0
= > Clearly 4 b² c² & 4 a² d² cancel out .
= > - 8 b c a d - 4 a² c² - b² d² = 0
= > - 4 ( a² c² + b² d² + 2 a c b d ) = 0
= > ( a c + b d )² = 0
= > a c + b d = 0
Hence proved.
Step-by-step explanation:
hope this helps please follow
Answer:
.e. D = 0
= > b² - 4 ac = 0
From question we have given :
b = 2 ( b c - a d )
a = a² + b²
c = c² - d²
Now put all value in condition.
= > ( 2 ( b c - a d )² - 4 * ( a² + b² ) ( c² + d² ) = 0
= > 4 b² c² +4 a² d² - 8 b c a d - 4 * ( a² c² + a² d²+ b² c² + b² d² ) = 0
= >4 b² c² +4 a² d² - 8 b c a d - 4 a² c² - 4 a² d² - 4 b² c² - b² d² = 0
= > Clearly 4 b² c² & 4 a² d² cancel out .
= > - 8 b c a d - 4 a² c² - b² d² = 0
= > - 4 ( a² c² + b² d² + 2 a c b d ) = 0
= > ( a c + b d )² = 0
= > a c + b d = 0
Hence proved.
Step-by-step explanation:
hope this helps please follow