Use the Quadratic Formula to solve for $y$ . $14 x ^ { 2 } - 10 x y + 9 y ^ { 2 } - 41 = 0$
A) $y = \frac { 14 x \pm \sqrt { 100 x ^ { 2 } + 36 \left( x ^ { 2 } - 9 \right) } } { - 18 }$
B) $y = \frac { 14 x \pm \sqrt { 100 x ^ { 2 } + 36 \left( x ^ { 2 } + 9 \right) } } { 18 }$
C) $y = \frac { 10 x \pm \sqrt { 100 x ^ { 2 } + 36 \left( 14 x ^ { 2 } - 41 \right) } } { - 18 }$
D) $y = \frac { 10 x \pm \sqrt { 100 x ^ { 2 } + 36 \left( 14 x ^ { 2 } + 41 \right) } } { 18 }$
E) $y = \frac { 10 x \pm \sqrt { 100 x ^ { 2 } - 36 \left( 14 x ^ { 2 } - 41 \right) } } { 18 }$

Question

Use the Quadratic Formula to solve for $y$ .  $14 x ^ { 2 } - 10 x y + 9 y ^ { 2 } - 41 = 0$ 
A) $y = \frac { 14 x \pm \sqrt { 100 x ^ { 2 } + 36 \left( x ^ { 2 } - 9 \right) } } { - 18 }$
B) $y = \frac { 14 x \pm \sqrt { 100 x ^ { 2 } + 36 \left( x ^ { 2 } + 9 \right) } } { 18 }$
C) $y = \frac { 10 x \pm \sqrt { 100 x ^ { 2 } + 36 \left( 14 x ^ { 2 } - 41 \right) } } { - 18 }$
D) $y = \frac { 10 x \pm \sqrt { 100 x ^ { 2 } + 36 \left( 14 x ^ { 2 } + 41 \right) } } { 18 }$
E) $y = \frac { 10 x \pm \sqrt { 100 x ^ { 2 } - 36 \left( 14 x ^ { 2 } - 41 \right) } } { 18 }$

Quizplus · Accepted Answer

The quadratic formula for $y$ in the equation $14x^2 - 10xy + 9y^2 - 41 = 0$ is derived by treating it as a quadratic in $y$: $ay^2 + by + c = 0$ with $a = 9$, $b = -10x$, and $c = 14x^2 - 41$. The formula is $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Substituting the values, we get $y = \frac{10x \pm \sqrt{100x^2 - 36(14x^2 - 41)}}{18}$.

Use the Quadratic Formula to Solve For yyy 14x2−10xy+9y2−41=014 x ^ { 2 } - 10 x y + 9 y ^ { 2 } - 41 = 014x2−10xy+9y2−41=0

Use the Quadratic Formula to Solve For $y$ $14 x ^ { 2 } - 10 x y + 9 y ^ { 2 } - 41 = 0$