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Find all real solutions \( (x, y) \) of the system. \( \begin{array}{l}x^{2}-y=4 \\ x^{2}+y^{2}=4\end{array} \)

Ask by Savage Bob. in the United States
Mar 20,2025

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The real solutions are \( (x, y) = (-2, 0) \), \( (-\sqrt{3}, -1) \), \( (2, 0) \), and \( (\sqrt{3}, -1) \).

Solución

Solve the system of equations \( x^{2}-y=4;x^{2}+y^{2}=4 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x^{2}-y=4\\x^{2}+y^{2}=4\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}y=-4+x^{2}\\x^{2}+y^{2}=4\end{array}\right.\) - step2: Substitute the value of \(y:\) \(x^{2}+\left(-4+x^{2}\right)^{2}=4\) - step3: Simplify: \(-7x^{2}+16+x^{4}=4\) - step4: Move the expression to the left side: \(-7x^{2}+16+x^{4}-4=0\) - step5: Subtract the numbers: \(-7x^{2}+12+x^{4}=0\) - step6: Factor the expression: \(\left(2-x\right)\left(2+x\right)\left(3-x^{2}\right)=0\) - step7: Separate into possible cases: \(2-x=0\cup 2+x=0\cup 3-x^{2}=0\) - step8: Solve the equation: \(x=2\cup x=-2\cup x=\sqrt{3}\cup x=-\sqrt{3}\) - step9: Rearrange the terms: \(\left\{ \begin{array}{l}x=2\\y=-4+x^{2}\end{array}\right.\cup \left\{ \begin{array}{l}x=-2\\y=-4+x^{2}\end{array}\right.\cup \left\{ \begin{array}{l}x=\sqrt{3}\\y=-4+x^{2}\end{array}\right.\cup \left\{ \begin{array}{l}x=-\sqrt{3}\\y=-4+x^{2}\end{array}\right.\) - step10: Calculate: \(\left\{ \begin{array}{l}x=2\\y=0\end{array}\right.\cup \left\{ \begin{array}{l}x=-2\\y=0\end{array}\right.\cup \left\{ \begin{array}{l}x=\sqrt{3}\\y=-1\end{array}\right.\cup \left\{ \begin{array}{l}x=-\sqrt{3}\\y=-1\end{array}\right.\) - step11: Calculate: \(\left\{ \begin{array}{l}x=-2\\y=0\end{array}\right.\cup \left\{ \begin{array}{l}x=-\sqrt{3}\\y=-1\end{array}\right.\cup \left\{ \begin{array}{l}x=2\\y=0\end{array}\right.\cup \left\{ \begin{array}{l}x=\sqrt{3}\\y=-1\end{array}\right.\) - step12: Check the solution: \(\left\{ \begin{array}{l}x=-2\\y=0\end{array}\right.\cup \left\{ \begin{array}{l}x=-\sqrt{3}\\y=-1\end{array}\right.\cup \left\{ \begin{array}{l}x=2\\y=0\end{array}\right.\cup \left\{ \begin{array}{l}x=\sqrt{3}\\y=-1\end{array}\right.\) - step13: Rewrite: \(\left(x,y\right) = \left(-2,0\right)\cup \left(x,y\right) = \left(-\sqrt{3},-1\right)\cup \left(x,y\right) = \left(2,0\right)\cup \left(x,y\right) = \left(\sqrt{3},-1\right)\) The real solutions of the system of equations are \( (x, y) = (-2, 0) \), \( (x, y) = (-\sqrt{3}, -1) \), \( (x, y) = (2, 0) \), and \( (x, y) = (\sqrt{3}, -1) \).

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The Deep Dive

To solve the system of equations given by \[ \begin{array}{l} x^{2} - y = 4 \\ x^{2} + y^{2} = 4 \end{array} \] we can start by expressing \( y \) from the first equation. Rearranging the first equation gives us: \[ y = x^2 - 4 \] Now we will substitute this expression for \( y \) into the second equation: \[ x^2 + (x^2 - 4)^2 = 4 \] Expanding \( (x^2 - 4)^2 \): \[ (x^2 - 4)^2 = x^4 - 8x^2 + 16 \] Substituting this back into the equation: \[ x^2 + x^4 - 8x^2 + 16 = 4 \] Combining like terms: \[ x^4 - 7x^2 + 16 - 4 = 0 \quad \Rightarrow \quad x^4 - 7x^2 + 12 = 0 \] Now, let \( z = x^2 \). The equation becomes a quadratic in terms of \( z \): \[ z^2 - 7z + 12 = 0 \] We can use the quadratic formula: \[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot 12}}{2 \cdot 1} \] Calculating the discriminant: \[ b^2 - 4ac = 49 - 48 = 1 \] Thus, we find: \[ z = \frac{7 \pm 1}{2} \] Calculating the two possible values for \( z \): 1. \( z = \frac{8}{2} = 4 \) 2. \( z = \frac{6}{2} = 3 \) Now, substituting back for \( x^2 \): 1. If \( x^2 = 4 \), then \( x = 2 \) or \( x = -2 \). 2. If \( x^2 = 3 \), then \( x = \sqrt{3} \) or \( x = -\sqrt{3} \). Next, we calculate \( y \) for each \( x \) using \( y = x^2 - 4 \): 1. For \( x = 2 \): \[ y = 2^2 - 4 = 0 \quad \Rightarrow \quad (2, 0) \] 2. For \( x = -2 \): \[ y = (-2)^2 - 4 = 0 \quad \Rightarrow \quad (-2, 0) \] 3. For \( x = \sqrt{3} \): \[ y = (\sqrt{3})^2 - 4 = 3 - 4 = -1 \quad \Rightarrow \quad (\sqrt{3}, -1) \] 4. For \( x = -\sqrt{3} \): \[ y = (-\sqrt{3})^2 - 4 = 3 - 4 = -1 \quad \Rightarrow \quad (-\sqrt{3}, -1) \] Now we can summarize the solutions: The real solutions \( (x, y) \) of the system are: \[ (2, 0), \, (-2, 0), \, (\sqrt{3}, -1), \, (-\sqrt{3}, -1) \]

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