Example Question - factoring quadratic equations
Here are examples of questions we've helped users solve.
Solving Quadratic Equations
The equation in the image is a quadratic equation. It reads: -m² - 7m - 4 = -2m². To solve for m, we should first simplify and rearrange the equation to standard quadratic form ax² + bx + c = 0. We can start by adding 2m² to both sides to combine like terms: -m² + 2m² - 7m - 4 = 0, which simplifies to: m² - 7m - 4 = 0. Now, we have a standard quadratic equation that we can solve by factoring, completing the square, or using the quadratic formula. This equation doesn't easily factor, so let's use the quadratic formula: m = [-b ± sqrt(b² - 4ac)] / (2a), where a = 1, b = -7, and c = -4. Plugging in these values: m = [-(-7) ± sqrt((-7)² - 4 * 1 * (-4))] / (2 * 1), m = [7 ± sqrt(49 + 16)] / 2, m = [7 ± sqrt(65)] / 2. So the solutions for m are: m = (7 + sqrt(65)) / 2 and m = (7 - sqrt(65)) / 2.
Solving Quadratic Equations Involving Square Roots
To solve the equation \( x - 1 - \sqrt{2x + 6} = 0 \), we can start by isolating the square root term: \( x - 1 = \sqrt{2x + 6} \) Next, we square both sides of the equation to eliminate the square root: \( (x - 1)^2 = (\sqrt{2x + 6})^2 \) \( x^2 - 2x + 1 = 2x + 6 \) Now let's bring all terms to one side of the equation: \( x^2 - 2x - 2x + 1 - 6 = 0 \) Which simplifies to: \( x^2 - 4x - 5 = 0 \) This is a quadratic equation that can be factored: \( (x - 5)(x + 1) = 0 \) Setting each factor equal to zero gives us the potential solutions: \( x - 5 = 0 \) or \( x + 1 = 0 \) So: \( x = 5 \) or \( x = -1 \) However, we have to check these solutions in the original equation because the squaring step can introduce extraneous solutions (solutions that do not actually work for the original equation). Let's check both values: When \( x = 5 \), substitute back into the original equation: \( 5 - 1 - \sqrt{2(5) + 6} = 0 \) \( 4 - \sqrt{10 + 6} = 0 \) \( 4 - \sqrt{16} = 0 \) \( 4 - 4 = 0 \) Which is true, so \( x = 5 \) is a solution. When \( x = -1 \), substitute back into the original equation: \( -1 - 1 - \sqrt{2(-1) + 6} = 0 \) \( -2 - \sqrt{-2 + 6} = 0 \) \( -2 - \sqrt{4} = 0 \) \( -2 - 2 = 0 \) Which is not true, so \( x = -1 \) is not a solution. Therefore, the solution to the equation is \( x = 5 \).
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