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Solve for all solutions of \( x \) \( x^{2}+6 x-15=0 \)

Ask by Sherman Hanson. in the United States
Feb 27,2025

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Tutor-Verified Answer

Answer

The solutions are \( x = -3 - 2\sqrt{6} \) and \( x = -3 + 2\sqrt{6} \).

Solution

Solve the equation \( x^{2}+6x-15=0 \). Solve the quadratic equation by following steps: - step0: Solve using the quadratic formula: \(x^{2}+6x-15=0\) - step1: Solve using the quadratic formula: \(x=\frac{-6\pm \sqrt{6^{2}-4\left(-15\right)}}{2}\) - step2: Simplify the expression: \(x=\frac{-6\pm \sqrt{96}}{2}\) - step3: Simplify the expression: \(x=\frac{-6\pm 4\sqrt{6}}{2}\) - step4: Separate into possible cases: \(\begin{align}&x=\frac{-6+4\sqrt{6}}{2}\\&x=\frac{-6-4\sqrt{6}}{2}\end{align}\) - step5: Simplify the expression: \(\begin{align}&x=-3+2\sqrt{6}\\&x=\frac{-6-4\sqrt{6}}{2}\end{align}\) - step6: Simplify the expression: \(\begin{align}&x=-3+2\sqrt{6}\\&x=-3-2\sqrt{6}\end{align}\) - step7: Rewrite: \(x_{1}=-3-2\sqrt{6},x_{2}=-3+2\sqrt{6}\) The solutions to the equation \(x^{2}+6x-15=0\) are \(x_{1}=-3-2\sqrt{6}\) and \(x_{2}=-3+2\sqrt{6}\).

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Beyond the Answer

To solve the quadratic equation \( x^{2}+6x-15=0 \), we can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1 \), \( b = 6 \), and \( c = -15 \). Plugging in these values: \[ x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-15)}}{2 \cdot 1} \] Calculating the discriminant: \[ 6^2 - 4 \cdot 1 \cdot (-15) = 36 + 60 = 96 \] Now substituting this into the formula: \[ x = \frac{-6 \pm \sqrt{96}}{2} \] Since \( \sqrt{96} = \sqrt{16 \cdot 6} = 4\sqrt{6} \), we have: \[ x = \frac{-6 \pm 4\sqrt{6}}{2} \] This simplifies to: \[ x = -3 \pm 2\sqrt{6} \] Therefore, the solutions for \( x \) are: \[ x = -3 + 2\sqrt{6} \quad \text{and} \quad x = -3 - 2\sqrt{6} \]

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