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Solve for \( x \) by first completing the squere

Ask by Horton Reid. in South Africa
Feb 03,2025

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To solve the equation \( x^2 + 6x + 5 = 0 \) by completing the square, follow these steps: 1. Move the constant term: \( x^2 + 6x = -5 \). 2. Find half of 6, which is 3, and square it to get 9. 3. Add 9 to both sides: \( x^2 + 6x + 9 = 4 \). 4. Write as a perfect square: \( (x + 3)^2 = 4 \). 5. Take the square root of both sides: \( x + 3 = \pm 2 \). 6. Solve for \( x \): \( x = -1 \) or \( x = -5 \). So, the solutions are \( x = -1 \) and \( x = -5 \).

Solución

Certainly! Completing the square is a useful method for solving quadratic equations of the form: \[ ax^2 + bx + c = 0 \] Let's walk through the process step-by-step with a specific example. Suppose we want to solve the quadratic equation: \[ x^2 + 6x + 5 = 0 \] **Step 1: Move the Constant Term to the Other Side** First, subtract the constant term from both sides to set up the equation for completing the square. \[ x^2 + 6x = -5 \] **Step 2: Complete the Square** To complete the square, take half of the coefficient of \( x \), square it, and add it to both sides of the equation. 1. **Find half of the coefficient of \( x \):** \[ \frac{6}{2} = 3 \] 2. **Square this value:** \[ 3^2 = 9 \] 3. **Add this square to both sides:** \[ x^2 + 6x + 9 = -5 + 9 \] **Step 3: Simplify Both Sides** Now, simplify both sides of the equation. \[ x^2 + 6x + 9 = 4 \] The left side of the equation is now a perfect square trinomial. **Step 4: Write the Perfect Square Trinomial as a Binomial Squared** \[ (x + 3)^2 = 4 \] **Step 5: Solve for \( x \)** Take the square root of both sides to solve for \( x \). \[ x + 3 = \pm \sqrt{4} \] \[ x + 3 = \pm 2 \] Now, isolate \( x \): 1. **First solution:** \[ x + 3 = 2 \] \[ x = 2 - 3 \] \[ x = -1 \] 2. **Second solution:** \[ x + 3 = -2 \] \[ x = -2 - 3 \] \[ x = -5 \] **Step 6: Write the Solutions** The solutions to the equation \( x^2 + 6x + 5 = 0 \) are: \[ x = -1 \quad \text{and} \quad x = -5 \] **Verification** To ensure that these solutions are correct, substitute them back into the original equation. 1. **For \( x = -1 \):** \[ (-1)^2 + 6(-1) + 5 = 1 - 6 + 5 = 0 \] 2. **For \( x = -5 \):** \[ (-5)^2 + 6(-5) + 5 = 25 - 30 + 5 = 0 \] Both solutions satisfy the original equation. --- **General Steps to Complete the Square:** If you have a general quadratic equation: \[ ax^2 + bx + c = 0 \] Follow these steps: 1. **Ensure the coefficient of \( x^2 \) is 1.** If not, divide the entire equation by \( a \). 2. **Move the constant term to the other side of the equation.** 3. **Take half of the coefficient of \( x \), square it, and add it to both sides.** 4. **Rewrite the left side as a perfect square trinomial.** 5. **Take the square root of both sides.** 6. **Solve for \( x \).** This method not only helps in solving quadratic equations but also in understanding the properties of parabolas and the derivation of the quadratic formula.

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Bonus Knowledge

Completing the square is like transforming a quadratic equation into a perfect square trinomial, which makes it easier to solve for \( x \). Take a standard quadratic form, \( ax^2 + bx + c = 0 \). First, move \( c \) to the other side: \( ax^2 + bx = -c \). Then, divide all terms by \( a \) if \( a \neq 1 \). Next, add \( \left(\frac{b}{2}\right)^2 \) to both sides, creating a perfect square on the left. For instance, if you have \( x^2 + 6x = -5 \), you would rewrite it as \( x^2 + 6x + 9 = 4 \) (adding 9, which is \( \left(\frac{6}{2}\right)^2 \)). This gives you \( (x + 3)^2 = 4 \). Taking the square root of both sides leads to \( x + 3 = ±2 \), yielding solutions \( x = -1 \) and \( x = -5 \). Voilà! Next steps often lead to confusion, but it’s all about isolating \( x \) carefully. A common mistake is forgetting to balance both sides of the equation after adding or subtracting, which can skew your solution. Always check that you apply operations equally on both sides to maintain equality. Happy solving!

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