Summary
Highlights
Quadratic equations can be solved using two main methods: factorization (making brackets) or the quadratic formula. The quadratic formula is reliable for any trinomial or expression where the highest exponent is two, even if it's easy to factorize.
For x^2 - 3x - 4 = 0, factorization leads to (x-4)(x+1)=0, resulting in x=4 or x=-1. The quadratic formula also produces these results, highlighting the importance of using brackets for negative 'b' values to avoid calculation errors.
When an equation contains x^2, always move all terms to one side to set the other side to zero. This is crucial for both factorization and applying the quadratic formula effectively.
For x^2 - 7x + 10 = 0, factorizing gives (x-5)(x-2)=0, leading to x=5 or x=2. This example reinforces the importance of setting the equation to zero.
For 2x^2 - 7x + 3 = 0, it's more efficient to directly use the quadratic formula rather than attempting to factorize in a test setting. The solutions are x=3 and x=0.5 (or 1/2).
For -x^2 + 11x - 8 = 0, apply the quadratic formula diligently with brackets. The answers, rounded to two decimal places, are x=0.78 and x=10.22.
For x^2 - 3x = 0, recognize it as a common factor problem, not a trinomial. Factor out x to get x(x-3)=0, leading to x=0 or x=3. The quadratic formula can also be used here by setting the 'c' value to zero.
Do not cancel x terms in x^2 = 4x. Instead, bring all terms to one side to get x^2 - 4x = 0, then factorize by common factor: x(x-4)=0. This yields x=0 and x=4, avoiding the loss of one solution.
For the final example, which is not easily factorizable, directly use the quadratic formula. The solutions are approximately x=2.3 and x=0.56.