Respuesta :
Answer:
Option A
The values of y are: [tex]y=-2+\sqrt{7}[/tex] or [tex]y=-2-\sqrt{7}[/tex].
Step-by-step explanation:
[tex]y^2+4y+4=7[/tex]
Subtract 7 from both the sides:
[tex]y^2+4y+4-7=7-7[/tex]
On Simplify:
[tex]y^2+4y-3=0[/tex]
Solve with the quadratic formula:
For a quadratic equation of the form [tex]ax^2+bx+c=0[/tex] the solutions are:
[tex]x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
For a=1, b=4 and c=-3 we have,
[tex]y_{1,2}=\frac{-4\pm \sqrt{4^2-4\cdot 1 \cdot (-3)}} {2\cdot 1}[/tex]
[tex]y_{1,2}=\frac{-4\pm\sqrt{16+12}} {2}[/tex]=[tex]\frac{-4\pm\sqrt{28}} {2}[/tex]
on solving we get, [tex]y_{1,2}=\frac{-4\pm2\sqrt{7}}{2}[/tex]=[tex]-2\pm \sqrt{7}[/tex]
therefore, the values of y are:
[tex]y=-2+\sqrt{7}[/tex] or [tex]y=-2-\sqrt{7}[/tex]
