It's a common sight in math textbooks and problem sets: the phrase 'solve 3x²'. At first glance, it might seem a bit cryptic, especially when you encounter variations like 'solve 3x² - 7x + 4 = 0' or 'solve 3x² + 2x - 13 = 0'. But really, it's just a shorthand for a fundamental mathematical task – finding the value(s) of 'x' that make a specific equation true.
Let's break down the simplest form, 'solve 3x²'. This is a very basic quadratic equation. To solve it, we're essentially asking, 'What number, when multiplied by itself and then by 3, gives us a certain result?' However, the query 'solve 3x²' on its own is incomplete. It's like saying 'find the number' without specifying what it should equal. To actually solve it, we need an equals sign and a value on the other side, such as 'solve 3x² = 27'.
When we see 'solve 3x² = 27', we're looking for the number(s) 'x' that, when squared and then multiplied by 3, result in 27. The process is straightforward: first, isolate the x² term by dividing both sides by 3. So, 3x² / 3 = 27 / 3, which simplifies to x² = 9. Now, we need to find the number(s) that, when squared, give us 9. This is where we take the square root of both sides. The square root of 9 is 3. But here's a crucial point in algebra: both positive and negative numbers, when squared, result in a positive number. So, both 3 * 3 and (-3) * (-3) equal 9. Therefore, the solutions to 'solve 3x² = 27' are x = 3 and x = -3.
Things get a bit more involved when we have a full quadratic equation, like 'solve 3x² - 7x + 4 = 0' or 'solve 3x² + 6x - 5 = 0'. These are equations in the standard form ax² + bx + c = 0. For these, we often turn to the quadratic formula, a powerful tool that gives us the solutions for 'x' regardless of the specific numbers. The formula is: x = [-b ± √(b² - 4ac)] / 2a.
Let's take 'solve 3x² + 6x - 5 = 0' as an example. Here, a = 3, b = 6, and c = -5. Plugging these into the formula:
x = [-6 ± √(6² - 4 * 3 * -5)] / (2 * 3) x = [-6 ± √(36 + 60)] / 6 x = [-6 ± √96] / 6
Now, √96 can be simplified, but for practical purposes, we often use a calculator. √96 is approximately 9.798. So, we have two possible solutions:
x₁ = (-6 + 9.798) / 6 ≈ 3.798 / 6 ≈ 0.633 x₂ = (-6 - 9.798) / 6 ≈ -15.798 / 6 ≈ -2.633
These are approximate solutions, often rounded to a certain number of significant figures, as seen in the reference material which gives 0.633 and -2.63.
Another variation, like 'solve 3x² + 2x - 13 = 0', might require answers in a specific format, such as 'a ± b√c'. For this, we again use the quadratic formula:
a = 3, b = 2, c = -13 x = [-2 ± √(2² - 4 * 3 * -13)] / (2 * 3) x = [-2 ± √(4 + 156)] / 6 x = [-2 ± √160] / 6
Here, √160 can be simplified. 160 is 16 * 10, so √160 = √16 * √10 = 4√10. Substituting this back:
x = [-2 ± 4√10] / 6
To get it into the 'a ± b√c' form, we divide each term in the numerator by the denominator (6):
x = -2/6 ± (4√10)/6 x = -1/3 ± (2/3)√10
So, the solutions are -1/3 + (2/3)√10 and -1/3 - (2/3)√10. It's fascinating how a simple 'solve 3x²' can lead to such diverse and interesting mathematical explorations, from basic square roots to the intricacies of the quadratic formula and surds.
Ultimately, 'solve 3x²' and its more complex relatives are all about finding the unknown. They're the building blocks for understanding how equations work and how we can use them to model and solve problems in the real world, whether it's calculating areas, understanding growth patterns, or even figuring out the trajectory of a projectile.
