Math, asked by mbansals2006, 19 days ago

find the roots of 2x² - 3(a+b)x + a² + 2ab + b² =0​

Answers

Answered by pratimakolkata9
1

Answer:

x = (a + b), (a + b) /2

Step-by-step explanation:

Given 2x² - 3(a+b)x + a² + 2ab + b² = 0

= 2x² - 3(a+b)x + (a + b) ² =0

=> x = [3(a + b) +, - sq( 9(a + b) ² - 8(a + b) ²)]/4

=> x = [3(a + b) +, - sq(a + b)²]/4

=> x = 2(a + b)/4, 4(a + b) /4

=> x = (a + b), (a + b) /2

Answered by eshaangupta1996
1

Answer:

Answer:

x = (a + b), (a + b) /2

Given 2x² - 3(a+b)x + a² + 2ab + b² = 0

= 2x² - 3(a+b)x + (a + b) ² =0

=> x = [3(a + b) +, - sq( 9(a + b) ² - 8(a + b) ²)]/4

=> x = [3(a + b) +, - sq(a + b)²]/4

=> x = 2(a + b)/4, 4(a + b) /4

=> x = (a + b), (a + b) /2

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