justinhendrix98 justinhendrix98
  • 03-11-2019
  • Mathematics
contestada

Solve the equation for all real solution 4y^2+19y+17=0

Respuesta :

funkeyusuff52
funkeyusuff52 funkeyusuff52
  • 03-11-2019

Answer:

y=-1.20 or -3.56

Step-by-step explanation:

4y^2+19y+17=0

subtract 7 from both sides to have

4y^2+19y=-17

Divide through by 4

y^2+19/4 y=-17/4

Complete the square of LHS to obtain

(y+19/8)^2 -(19/8)^2=-17/4

simplifying further we have

(y+19/8)^2=-17/4+(19)^2 /(8)^2 =-17/4+361/64=-272+361/64=89/64

i.e. (y+19/8)^2=89/64

taking the square root of both sides, we obtain

y+19/8=sqrt(89/64)=+ or - 1.18

which implies y= -19/8+or-1.18=2.38+or-1.18

either y = -2.38+1.18=-1.2

or y= -2.38-1.18= -3.56

Therefore, y=-1.20 or -3.56

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