Use the Quadratic Formula to solve for $y$ . $x ^ { 2 } + x y + 4 y ^ { 2 } + x + y - 4 = 0$
A) $y = \frac { ( x + 1 ) \pm \sqrt { ( x - 1 ) ^ { 2 } - 8 \left( x ^ { 2 } - x + 4 \right) } } { 4 }$
B) $y = \frac { - ( x + 1 ) \pm \sqrt { ( x + 1 ) ^ { 2 } - 8 \left( x ^ { 2 } + x + 4 \right) } } { 4 }$
C) $y = \frac { - ( x + 1 ) \pm \sqrt { ( x + 1 ) ^ { 2 } - 8 \left( x ^ { 2 } + x - 4 \right) } } { 4 }$
D) $y = \frac { ( x + 1 ) \pm \sqrt { ( x + 1 ) ^ { 2 } - 8 \left( x ^ { 2 } + x - 4 \right) } } { 4 }$
E) $y = \frac { - ( x + 1 ) \pm \sqrt { ( x + 1 ) ^ { 2 } + 8 \left( x ^ { 2 } + x - 4 \right) } } { 4 }$

Question

Use the Quadratic Formula to solve for $y$ .  $x ^ { 2 } + x y + 4 y ^ { 2 } + x + y - 4 = 0$ 
A) $y = \frac { ( x + 1 ) \pm \sqrt { ( x - 1 ) ^ { 2 } - 8 \left( x ^ { 2 } - x + 4 \right) } } { 4 }$
B) $y = \frac { - ( x + 1 ) \pm \sqrt { ( x + 1 ) ^ { 2 } - 8 \left( x ^ { 2 } + x + 4 \right) } } { 4 }$
C) $y = \frac { - ( x + 1 ) \pm \sqrt { ( x + 1 ) ^ { 2 } - 8 \left( x ^ { 2 } + x - 4 \right) } } { 4 }$
D) $y = \frac { ( x + 1 ) \pm \sqrt { ( x + 1 ) ^ { 2 } - 8 \left( x ^ { 2 } + x - 4 \right) } } { 4 }$
E) $y = \frac { - ( x + 1 ) \pm \sqrt { ( x + 1 ) ^ { 2 } + 8 \left( x ^ { 2 } + x - 4 \right) } } { 4 }$

Quizplus · Accepted Answer

The given equation is a quadratic equation in terms of $y$. To solve for $y$, we need to rearrange the equation in the form of $ay^2 + by + c = 0$, where $a$, $b$, and $c$ are coefficients that may depend on $x$. The given equation is $x^2 + xy + 4y^2 + x + y - 4 = 0$. Grouping terms gives us $4y^2 + (x + 1)y + (x^2 + x - 4) = 0$. Here, $a = 4$, $b = x + 1$, and $c = x^2 + x - 4$. Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we substitute $a$, $b$, and $c$ to get $y = \frac{-(x + 1) \pm \sqrt{(x + 1)^2 - 4*4*(x^2 + x - 4)}}{2*4}$, which simplifies to $y = \frac{-(x + 1) \pm \sqrt{(x + 1)^2 - 8(x^2 + x - 4)}}{4}$. This matches option C.

Use the Quadratic Formula to Solve For yyy x2+xy+4y2+x+y−4=0x ^ { 2 } + x y + 4 y ^ { 2 } + x + y - 4 = 0x2+xy+4y2+x+y−4=0

Use the Quadratic Formula to Solve For $y$ $x ^ { 2 } + x y + 4 y ^ { 2 } + x + y - 4 = 0$