Solve the equation for the indicated variable. Assume no denominator is 0.
-$2 x ^ { 2 } - 4 x y + 3 y ^ { 2 } = 1$, for $x$
A) $x = - y \pm 2 \sqrt { 1 - y ^ { 2 } }$
B) $x = \frac { 2 y \pm \sqrt { 2 - 2 y ^ { 2 } } } { 2 }$
C) $x = y \pm ( 1 - y )$
D) $x = - y \pm \sqrt { 1 - y ^ { 2 } }$

Question

Quizplus · Accepted Answer

To solve for $x$, we rearrange the equation into the form of a quadratic equation $ax^2 + bx + c = 0$. The given equation is $2x^2 - 4xy + 3y^2 = 1$. We want to isolate $x$, so we move everything except the terms with $x$ to the other side: $2x^2 - 4xy = 1 - 3y^2$. This equation can be rewritten as $2x^2 - 4xy + 3y^2 - 3y^2 = 1 - 3y^2$, simplifying to $2x^2 - 4xy = 1 - 3y^2$. To solve for $x$, we use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 2$, $b = -4y$, and $c = -(1 - 3y^2)$. Plugging these into the formula gives $x = \frac{4y \pm \sqrt{(-4y)^2 - 4(2)(-(1 - 3y^2))}}{2(2)}$, which simplifies to $x = \frac{4y \pm \sqrt{16y^2 + 8 - 24y^2}}{4}$, and further to $x = \frac{4y \pm \sqrt{-8y^2 + 8}}{4}$. Simplifying inside the square root gives $x = \frac{4y \pm \sqrt{8(1 - y^2)}}{4}$, and taking $8$ out of the square root as $2\sqrt{2}$ gives $x = \frac{4y \pm 2\sqrt{2}\sqrt{1 - y^2}}{4}$, which simplifies to $x = \frac{2y \pm \sqrt{2 - 2y^2}}{2}$.

Solve the Equation for the Indicated Variable 2x2−4xy+3y2=12 x ^ { 2 } - 4 x y + 3 y ^ { 2 } = 12x2−4xy+3y2=1

Solve the Equation for the Indicated Variable $2 x ^ { 2 } - 4 x y + 3 y ^ { 2 } = 1$