Example Question - factoring quadratic equation
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Solving a Quadratic Equation by Factoring
The equation provided in the image is a quadratic equation that can be solved by factoring, completing the square, or using the quadratic formula. Since the coefficients are quite large, I'll solve it by factoring for simplicity. The equation is: \[ 28x^2 - 72x - 180 = 0 \] To make it easier to factor, we can look for a common factor between the coefficients, which in this case is 4. Let's divide the whole equation by 4 to simplify: \[ 7x^2 - 18x - 45 = 0 \] Now, we look to factor the quadratic. We need to find two numbers that multiply to \(7 \times (-45)\) (which is -315) and at the same time add up to -18 (the coefficient of the x term). These numbers are -25 and +7. So, we can express the middle term -18x as -25x + 7x, like this: \[ 7x^2 - 25x + 7x - 45 = 0 \] Group the terms to factor by grouping: \[ (7x^2 - 25x) + (7x - 45) = 0 \] Factor out the common factors: \[ x(7x - 25) + 5(7x - 25) = 0 \] Now you can see that \( (7x - 25) \) is a common factor: \[ (x + 5)(7x - 25) = 0 \] To find the solutions for x, set each factor equal to zero: \[ x + 5 = 0 \quad \text{or} \quad 7x - 25 = 0 \] Solve both equations: \[ x = -5 \quad \text{or} \quad 7x = 25 \] \[ \quad \text{or} \quad x = \frac{25}{7} \approx 3.571 \] So, the solutions to the quadratic equation are: \[ x = -5 \quad \text{or} \quad x = \frac{25}{7} \]
Solving Quadratic Equation by Factoring
The equation given in the image is: x(x - 1) - 12 = 0 To solve this equation, let's first expand the brackets and then group terms: x^2 - x - 12 = 0 This is a quadratic equation of the form ax^2 + bx + c = 0, with a = 1, b = -1, and c = -12. We can solve this equation by factoring, completing the square, or using the quadratic formula. Since this equation can be factored easily, let's try factoring: We need two numbers that multiply to -12 and add up to -1. Those numbers are -4 and 3. So we can write the equation as: (x - 4)(x + 3) = 0 Now, set each factor equal to zero and solve for x: x - 4 = 0 or x + 3 = 0 x = 4 x = -3 Thus, the solutions to the equation are x = 4 and x = -3.
Solving a Quadratic Equation by Factoring
I see an equation written on a piece of paper: \[ g = 5x^2 - 10x + 5 \] To solve this equation for \( x \), first, let's see if we can factor it. A quadratic equation in the form \( ax^2 + bx + c = 0 \) can sometimes be factored into the form: \[ (dx + e)(fx + h) = 0 \] The given equation \( g = 5x^2 - 10x + 5 \) is not set equal to zero, so we can't solve for \( x \) until we do so. Let's assume you meant to solve for \( x \) when \( g = 0 \). That would give us: \[ 0 = 5x^2 - 10x + 5 \] Now, we clearly see that each term is divisible by 5. Let's factor out a 5: \[ 0 = 5(x^2 - 2x + 1) \] The quadratic inside the parentheses is a perfect square trinomial and factors as: \[ 0 = 5(x - 1)^2 \] Now we can find the value of \( x \) that makes the equation true by setting \( (x - 1)^2 \) equal to zero: \[ (x - 1)^2 = 0 \] Taking the square root of both sides gives us: \[ x - 1 = 0 \] So the solution for \( x \) is: \[ x = 1 \] Thus, the equation \( 5x^2 - 10x + 5 = 0 \) has a repeated real solution, which is \( x = 1 \).
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