Use the quadratic formula to solve the equation for x in terms of y and y in terms of x. (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION.)
2x^2 + 7xy + 1 − y^2 = 0
(a) x in terms of y
x =
(b) y in terms of x
y =
To solve the equation 2x^2 + 7xy + 1 − y^2 = 0 using the quadratic formula, we treat it as a quadratic equation in terms of either x or y.
(a) To find x in terms of y, we treat it as a quadratic equation in x. Rearrange the equation to isolate x:
2x^2 + 7xy + 1 − y^2 = 0
2x^2 + 7xy = y^2 - 1
2x^2 + 7xy = (y + 1)(y - 1)
Now we can apply the quadratic formula, which states that for a quadratic equation of the form ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = 2, b = 7y, and c = -(y + 1)(y - 1). Plugging these values into the quadratic formula, we get:
x = (-7y ± √((7y)^2 - 4(2)(-(y + 1)(y - 1)))) / (2(2))
Simplifying further:
x = (-7y ± √(49y^2 + 8(y + 1)(y - 1))) / 4
Therefore, x in terms of y is:
x = (-7y ± √(49y^2 + 8(y + 1)(y - 1))) / 4
(b) To find y in terms of x, we treat it as a quadratic equation in y. Rearrange the equation to isolate y:
2x^2 + 7xy + 1 − y^2 = 0
-y^2 + 7xy + 2x^2 + 1 = 0
Now we can apply the quadratic formula. In this case, a = -1, b = 7x, and c = 2x^2 + 1. Plugging these values into the quadratic formula, we get:
y = (-7x ± √((7x)^2 - 4(-1)(2x^2 + 1))) / (2(-1))
Simplifying further:
y = (-7x ± √(49x^2 + 8x^2 + 8)) / -2
Therefore, y in terms of x is:
y = (-7x ± √(57x^2 + 8)) / -2
In summary:
(a) x in terms of y: x = (-7y ± √(49y^2 + 8(y + 1)(y - 1))) / 4
(b) y in terms of x: y = (-7x ± √(57x^2 + 8)) / -2