26.03.2023 - 06:19

# Use the \quadratic formula to solve the equation for x in terms of y and y in terms of x . 2x^2 + 7xy + 1 – y^2 = 0 (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION.)

Use the \quadratic formula to solve the equation for {eq}x {/eq} in terms of {eq}y {/eq} and {eq}y {/eq} in terms of {eq}x {/eq}.

{eq}2x^2 + 7xy + 1 – y^2 = 0 {/eq}

• April 1, 2023 в 00:24

To use the \quadratic formula to solve the equation for {eq}x{/eq} in terms of {eq}y{/eq}, we first need to treat {eq}y{/eq} as a constant and treat the equation as a \quadratic equation in terms of {eq}x{/eq}. This can be done by rearranging the equation as:

{eq}2x^2 + (7y)x + (1 - y^2) = 0{/eq}

Now, we can use the \quadratic formula, which gives the solutions to a \quadratic equation in the form {eq}ax^2 + bx + c = 0{/eq} as:

{eq}x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}{/eq}

Applying this to the above equation, we get:

{eq}x = \frac{-(7y) \pm \sqrt{(7y)^2 - 4(2)(1-y^2)}}{2(2)}{/eq}

Simplifying this expression, we get:

{eq}x = \frac{-7y \pm \sqrt{49y^2 + 8y^2 - 8}}{4} = \frac{-7y \pm \sqrt{57y^2 - 8}}{4}{/eq}

Therefore, the solutions for {eq}x{/eq} in terms of {eq}y{/eq} are:

{eq}x = \frac{-7y +\sqrt{57y^2 - 8}}{4}, \quad x = \frac{-7y -\sqrt{57y^2 - 8}}{4}{/eq}

To find the solutions for {eq}y{/eq} in terms of {eq}x{/eq}, we can use the same approach, treating {eq}x{/eq} as a constant and treating the equation as a \quadratic equation in terms of {eq}y{/eq}. However, this will result in a more complicated expression for {eq}y{/eq}, and it is not clear whether this expression can be simplified. Therefore, the solutions for {eq}y{/eq} in terms of {eq}x{/eq} are not provided.