Solving Quadratic Equations: Find X In 2x² - 9x + 9 = 0

Hey everyone! Today, let's tackle a classic quadratic equation. We're going to solve for x in the equation $2x^2 - 9x + 9 = 0$. Quadratic equations pop up everywhere, from physics problems to computer graphics, so mastering them is super useful. We'll break down the steps and make sure you understand how to find both solutions.

Understanding Quadratic Equations

First off, let's understand what we're dealing with. A quadratic equation is basically any equation that can be written in the form $ax^2 + bx + c = 0$, where a, b, and c are constants, and a isn't zero (otherwise, it's just a linear equation!). The solutions to a quadratic equation are also called its roots or zeros, and they're the values of x that make the equation true.

In our case, we have $2x^2 - 9x + 9 = 0$. So, a = 2, b = -9, and c = 9. There are a few ways to solve quadratic equations, including factoring, completing the square, and using the quadratic formula. For this particular equation, factoring is a pretty straightforward approach. Factoring involves rewriting the quadratic expression as a product of two binomials. When the product of these two binomials equals zero, at least one of them must be zero, which gives us our solutions for x.

Factoring the Quadratic Equation

Okay, so how do we factor $2x^2 - 9x + 9$? This can sometimes feel like a puzzle, but here’s the general idea. We're looking for two binomials of the form $(px + q)(rx + s)$ such that when we multiply them out, we get $2x^2 - 9x + 9$. Basically, we need to find p, q, r, and s so that the equation holds true.

The first term $2x^2$ tells us that p and r must multiply to 2. The only integer factors of 2 are 1 and 2, so we can start with $(2x + q)(x + s)$. Now we need to figure out q and s. We know that q times s must equal 9 (our constant term), and that $(2x * s) + (q * x)$ must equal -9x. So, we’re looking for two numbers that multiply to 9 and, when combined in the right way with the 2 from the 2x term, give us -9.

The factors of 9 are 1, 3, and 9. After a little trial and error (or a lot, depending on how your brain works!), we find that -3 and -3 work perfectly. Let’s try $(2x - 3)(x - 3)$. Expanding this, we get $2x^2 - 6x - 3x + 9 = 2x^2 - 9x + 9$. Bingo! That's exactly what we wanted.

Finding the Solutions

Now that we have factored the equation as $(2x - 3)(x - 3) = 0$, we can find the solutions for x. Remember, if the product of two things is zero, then at least one of them must be zero. So, either $2x - 3 = 0$ or $x - 3 = 0$.

Let's solve each of these equations separately:

1. $2x - 3 = 0$

   Add 3 to both sides: $2x = 3$

   Divide by 2: $x = \frac{3}{2}$

2. $x - 3 = 0$

   Add 3 to both sides: $x = 3$

So, our two solutions are x = $\frac{3}{2}$ and x = 3. These are the values of x that make the original equation $2x^2 - 9x + 9 = 0$ true. You can plug them back into the original equation to check if they work.

Checking the Solutions

It's always a good idea to check your answers, just to be sure. Let's plug each solution back into the original equation $2x^2 - 9x + 9 = 0$.

For $x = \frac{3}{2}$:

$2(\frac{3}{2})^2 - 9(\frac{3}{2}) + 9 = 2(\frac{9}{4}) - \frac{27}{2} + 9 = \frac{9}{2} - \frac{27}{2} + \frac{18}{2} = \frac{9 - 27 + 18}{2} = \frac{0}{2} = 0$. This checks out!

For $x = 3$:

$2(3)^2 - 9(3) + 9 = 2(9) - 27 + 9 = 18 - 27 + 9 = 0$. This one checks out too!

Both solutions work, so we can be confident in our answer.

Alternative Method: Quadratic Formula

Just for completeness, let’s also talk about the quadratic formula. If you can’t easily factor a quadratic equation, the quadratic formula is your best friend. It says that for any quadratic equation $ax^2 + bx + c = 0$, the solutions for x are given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In our case, a = 2, b = -9, and c = 9. Plugging these values into the quadratic formula, we get:

$x = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(2)(9)}}{2(2)} = \frac{9 \pm \sqrt{81 - 72}}{4} = \frac{9 \pm \sqrt{9}}{4} = \frac{9 \pm 3}{4}$

So, we have two possible solutions:

$x = \frac{9 + 3}{4} = \frac{12}{4} = 3$

$x = \frac{9 - 3}{4} = \frac{6}{4} = \frac{3}{2}$

As you can see, we get the same solutions as we did by factoring: x = 3 and x = $\frac{3}{2}$. The quadratic formula is a powerful tool that always works, even when factoring is difficult or impossible.

Key Takeaways

• Quadratic equations are of the form $ax^2 + bx + c = 0$.
• You can solve them by factoring, completing the square, or using the quadratic formula.
• The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Always check your solutions by plugging them back into the original equation.

Conclusion

So, there you have it! We've successfully solved the quadratic equation $2x^2 - 9x + 9 = 0$ and found that the solutions are x = $\frac{3}{2}$ and x = 3. Whether you prefer factoring or using the quadratic formula, understanding how to solve these equations is a valuable skill. Keep practicing, and you'll become a quadratic equation-solving pro in no time! Remember practice makes perfect so keep at it.

I hope this explanation helped you guys! Let me know if you have any questions or want to explore more math problems. Happy solving!