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Solve the equation \[ x^{2}=5 x+14 \]

Ask by Murray Schmidt. in the United States
Jan 24,2025

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Answer

The solutions to the equation \( x^{2} = 5x + 14 \) are \( x = 7 \) and \( x = -2 \).

Solution

To solve the quadratic equation: \[ x^{2} = 5x + 14 \] follow these steps: ### 1. Rewrite the Equation in Standard Form First, bring all terms to one side of the equation to set it to zero: \[ x^{2} - 5x - 14 = 0 \] This is the standard form of a quadratic equation: \[ ax^{2} + bx + c = 0 \] where: - \( a = 1 \) - \( b = -5 \) - \( c = -14 \) ### 2. Use the Quadratic Formula The quadratic formula provides the solutions to any quadratic equation: \[ x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \] Plugging in the values: \[ x = \frac{-(-5) \pm \sqrt{(-5)^{2} - 4(1)(-14)}}{2(1)} \] \[ x = \frac{5 \pm \sqrt{25 + 56}}{2} \] \[ x = \frac{5 \pm \sqrt{81}}{2} \] \[ x = \frac{5 \pm 9}{2} \] ### 3. Solve for the Two Possible Values of \( x \) 1. **First Solution (\(+\) sign):** \[ x = \frac{5 + 9}{2} = \frac{14}{2} = 7 \] 2. **Second Solution (\(-\) sign):** \[ x = \frac{5 - 9}{2} = \frac{-4}{2} = -2 \] ### 4. Conclusion The solutions to the equation \( x^{2} = 5x + 14 \) are: \[ x = 7 \quad \text{and} \quad x = -2 \] ### Verification To ensure the solutions are correct, substitute them back into the original equation. 1. **For \( x = 7 \):** \[ 7^{2} = 5(7) + 14 \implies 49 = 35 + 14 \implies 49 = 49 \quad \text{✓} \] 2. **For \( x = -2 \):** \[ (-2)^{2} = 5(-2) + 14 \implies 4 = -10 + 14 \implies 4 = 4 \quad \text{✓} \] Both solutions satisfy the original equation.

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To solve the equation \( x^{2} = 5x + 14 \), we first rearrange it to standard quadratic form: \[ x^{2} - 5x - 14 = 0 \] Next, we can either factor the equation, use the quadratic formula, or complete the square. Here, let's apply the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In our equation, \( a = 1 \), \( b = -5 \), and \( c = -14 \). Plugging these values into the formula gives: \[ x = \frac{-(-5) \pm \sqrt{(-5)^{2} - 4 \cdot 1 \cdot (-14)}}{2 \cdot 1} \] Calculating the discriminant: \[ b^2 - 4ac = 25 + 56 = 81 \] So we have: \[ x = \frac{5 \pm \sqrt{81}}{2} \] Since \( \sqrt{81} = 9 \), our equation becomes: \[ x = \frac{5 \pm 9}{2} \] This leads to two potential solutions: 1. \( x = \frac{5 + 9}{2} = \frac{14}{2} = 7 \) 2. \( x = \frac{5 - 9}{2} = \frac{-4}{2} = -2 \) Therefore, the solutions to the equation are: \[ x = 7 \quad \text{and} \quad x = -2 \]

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