Question

the value of y which will satisfy the equations 2x^2+6x+5y+1=0 and 2x+y+3=0 may be found by solving ​

Answer 1

Answer:

y = -5 ± 4√2

Step-by-step explanation:

the value of y which will satisfy the equations 2x^2+6x+5y+1=0 and 2x+y+3=0 may be found by solving

2x²+6x+5y+1=0  - eq 1

2x+y+3=0

x = -(y + 3)/2

Putting value of x in eq 1

2(-(y+3)/2)² + 6(-(y+3)/2) + 5y + 1 = 0

=> (2/4)( y² + 6y + 9) - 3(y+3) + 5y + 1 = 0

multiplying by 2both sides

=> y² + 6y + 9 - 6(y+3)  + 10y + 2 = 0

=> y² - 6y + 6y + 10y + 9 - 18 + 2 = 0

=> y² + 10y -7 = 0

y = ( -10 ± √(10² - 4(1)(-7)) )/2

=> y =  ( -10 ± √(100 + 28) )/2

=> y =  ( -10 ± √128 )/2

=> y =  ( -10 ± 8√2 )/2

=> y = -5 ± 4√2