Find the value or values of y in the quadratic equation y2 + 4y + 4 = 7. A. y = −2 + √ 7 , y = −2 − √ 7 B. y = 2, y = 7 C. y = −2√ 7 , y = 2√ 7 D. y = 2 + 2√ 7 , y = 2 − 2√ 7
Answer:
Option A
The values of y are: \(y=-2+\sqrt{7}\) or \(y=-2-\sqrt{7}\).
Step-by-step explanation:
\(y^2+4y+4=7\)
Subtract 7 from both the sides:
\(y^2+4y+4-7=7-7\)
On Simplify:
\(y^2+4y-3=0\)
Solve with the quadratic formula:
For a quadratic equation of the form \(ax^2+bx+c=0\) the solutions are:
\(x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)
For a=1, b=4 and c=-3 we have,
\(y_{1,2}=\frac{-4\pm \sqrt{4^2-4\cdot 1 \cdot (-3)}} {2\cdot 1}\)
\(y_{1,2}=\frac{-4\pm\sqrt{16+12}} {2}\)=\(\frac{-4\pm\sqrt{28}} {2}\)
on solving we get, \(y_{1,2}=\frac{-4\pm2\sqrt{7}}{2}\)=\(-2\pm \sqrt{7}\)
therefore, the values of y are:
\(y=-2+\sqrt{7}\) or \(y=-2-\sqrt{7}\)