Quadratic equations 1 Solve \( x^{2}-x-30=0 \) using the factorisation method. \( x^{2}-x-30=0 \) \( x^{2}+\square x-\square x-30=0 \) \( (x+\square)(x-\square)=0 \) \( x+\square-0 \) or \( x-\square= \) \( x=-\square \) or \( x=\square \)

To solve the equation \( x^{2} - x - 30 = 0 \) using the factorization method, we first look for two numbers that multiply to -30 and add up to -1. Those numbers are 5 and -6. So we rewrite the equation as \( x^{2} + 5x - 6x - 30 = 0 \). This can be factored into \( (x + 5)(x - 6) = 0 \). From here, we set each factor equal to zero: 1. \( x + 5 = 0 \) gives \( x = -5 \) 2. \( x - 6 = 0 \) gives \( x = 6 \) So the solutions are \( x = -5 \) and \( x = 6 \). Now that you've got the roots, the relationship between quadratic equations and their solutions is fascinating! Historically, these equations have been studied since ancient times, with various cultures documenting methods to solve them. The general quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), was refined through contributions from mathematicians across the world, from the Babylonians to the Indian and Islamic mathematicians! In practical scenarios, quadratic equations appear in a variety of real-world applications, such as physics (calculating projectile motion), economics (maximizing profit), or even in architecture (designing arches). If you’re focused on maximizing efficiency, for instance, you might set up a quadratic to find the optimal dimensions of a product or space!