Quadratic Formula Solved: $5x^2 - 8x + 5 = 0$

Jan 25, 2026 by Andrew McMorgan 46 views

Hey guys! Today, we're diving deep into the fascinating world of quadratic equations, specifically tackling one that requires the power of the quadratic formula. You know, those equations that look a little something like $ax^2 + bx + c = 0$ ? Well, the one we've got our sights set on is $5x^2 - 8x + 5 = 0$ . Our mission, should we choose to accept it (and we totally do!), is to find the solutions for $x$ and express them in a super specific format: $x= rac{ r - s i }{t}$ and $x= rac{ r + s i }{t}$ , where $r$ , $s$ , and $t$ are integers, and any fractions are simplified to their bare essence. This is gonna be a fun ride, so buckle up!

Understanding the Quadratic Formula: Your Mathematical Lifesaver

Alright, let's kick things off by making sure we're all on the same page about the star of the show: the quadratic formula. This bad boy is your go-to tool when you're faced with a quadratic equation that just won't factor nicely. Remember that standard form, $ax^2 + bx + c = 0$ ? The quadratic formula is derived from it and gives us the solutions for $x$ directly. It looks like this: $x = rac{-b eq oxed{\sqrt{b^2-4ac}}}{2a}$ . See that $eq$ symbol in there? That's actually a plus-minus sign, $f{oldsymbol{\pm}}$ . It signifies that there are usually two potential solutions for $x$ . The part under the square root, $b^2 - 4ac$ , is called the discriminant. It's a super important clue about the nature of our solutions. If the discriminant is positive, we get two distinct real solutions. If it's zero, we have exactly one real solution (a repeated root). And if it's negative, like in our case, we're headed into the territory of complex numbers, which is where we'll find that sneaky ' $i$ ' – the imaginary unit, where $i = oxed{\sqrt{-1}}$ . So, before we plug and chug, let's identify our $a$ , $b$ , and $c$ values from our specific equation, $5x^2 - 8x + 5 = 0$ . In this equation, $a = 5$ (the coefficient of $x^2$ ), $b = -8$ (the coefficient of $x$ ), and $c = 5$ (the constant term). Got it? Awesome! Now we're armed with the knowledge and the specific values needed to conquer this problem using our trusty quadratic formula. It’s all about breaking it down step-by-step, and the formula is our guide.

Plugging in the Values: Let the Calculation Begin!

Now that we've identified our coefficients $a=5$ , $b=-8$ , and $c=5$ , it's time to plug them into the quadratic formula: $x = rac{-b eq oxed{\sqrt{b^2-4ac}}}{2a}$ . Let's substitute carefully, paying close attention to those signs, guys. So, we have: $x = rac{-(-8) eq oxed{\sqrt{(-8)^2-4(5)(5)}}}{2(5)}$ . First off, the $-(-8)$ at the front simplifies to a positive $8$ . Next, let's tackle the discriminant inside the square root: $(-8)^2$ is $64$ , and $4(5)(5)$ is $4(25)$ , which equals $100$ . So, the discriminant becomes $64 - 100$ . This gives us $-36$ . And down below, in the denominator, $2(5)$ is simply $10$ . Putting it all together, our equation now looks like: $x = rac{8 eq oxed{\sqrt{-36}}}{10}$ . This is where things get interesting! Since we have a negative number under the square root, we know we're dealing with imaginary numbers. Remember, $oxed{\sqrt{-1}} = i$ . So, $oxed{\sqrt{-36}}$ can be rewritten as $oxed{\sqrt{36 imes -1}}$ , which is $oxed{\sqrt{36}} imes oxed{\sqrt{-1}}$ , or $6i$ . Substituting this back into our equation, we get: $x = rac{8 eq 6i}{10}$ . We're so close to the finish line now, just a few more steps to simplify this expression into the required format. It's all about careful substitution and understanding how the imaginary unit ' $i$ ' works its magic.

Simplifying to the Target Form: The Final Countdown

We've arrived at $x = rac{8 eq 6i}{10}$ , and our goal is to express this in the form $x= rac{ r - s i }{t}$ and $x= rac{ r + s i }{t}$ , where $r$ , $s$ , and $t$ are integers and fractions are simplified. To do this, we need to separate the real and imaginary parts and then simplify the entire fraction. First, let's address the $eq$ sign, which represents both a plus and a minus. This means we have two separate solutions:

$x_1 = rac{8 + 6i}{10}$
$x_2 = rac{8 - 6i}{10}$

Now, let's simplify each of these. We can see that the numerator and the denominator in both expressions share a common factor of $2$ . So, we divide both the $8$ , the $6i$ (or $-6i$ ), and the $10$ by $2$ .

For $x_1$ : $x_1 = rac{8 ext{ extdiv} 2 + 6i ext{ extdiv} 2}{10 ext{ extdiv} 2} = rac{4 + 3i}{5}$ For $x_2$ : $x_2 = rac{8 ext{ extdiv} 2 - 6i ext{ extdiv} 2}{10 ext{ extdiv} 2} = rac{4 - 3i}{5}$

These solutions are now in the form $rac{r eq si}{t}$ ! Let's match them up with the required format: $x= rac{ r - s i }{t}$ and $x= rac{ r + s i }{t}$ .

Our first solution, $x_1 = rac{4 + 3i}{5}$ , can be written as $x= rac{4 - (-3)i}{5}$ . Here, $r=4$ , $s=-3$ , and $t=5$ . However, the problem asks for $s$ to be a positive integer representing the magnitude of the imaginary part when written as $r+si$ . Let's stick to the standard interpretation where $s$ is positive and the sign is handled by the $eq$ or by how we write the final form.

Let's rewrite our simplified solutions to perfectly match the requested format. We have $rac{4+3i}{5}$ and $rac{4-3i}{5}$ .

If we write $x= rac{ r - s i }{t}$ , we can use the second solution: $x = rac{4 - 3i}{5}$ . In this case, $r=4$ , $s=3$ , and $t=5$ . These are all integers, and the fraction is simplified.

If we write $x= rac{ r + s i }{t}$ , we can use the first solution: $x = rac{4 + 3i}{5}$ . In this case, $r=4$ , $s=3$ , and $t=5$ . Again, these are all integers, and the fraction is simplified.

So, the two solutions, written in the specified form, are:

$x = rac{4 - 3i}{5}$ $x = rac{4 + 3i}{5}$

And in terms of $r, s, t$ as integers, we have $r=4$ , $s=3$ , and $t=5$ . We successfully navigated the quadratic formula, handled imaginary numbers, and simplified our results to the exact format required. High fives all around!

Conclusion: Mastering Quadratic Equations

So there you have it, guys! We've successfully solved the quadratic equation $5x^2 - 8x + 5 = 0$ using the quadratic formula. Remember, the key steps were identifying $a$ , $b$ , and $c$ ; carefully plugging these values into the formula $x = rac{-b eq oxed{\sqrt{b^2-4ac}}}{2a}$ ; simplifying the expression, especially when dealing with the discriminant ( $b^2-4ac$ ); and finally, expressing the solutions in the required format $x= rac{ r - s i }{t}$ and $x= rac{ r + s i }{t}$ . The process might seem a bit daunting at first, especially when imaginary numbers crop up, but by breaking it down into manageable steps, it becomes totally conquerable. The discriminant, $b^2 - 4ac$ , is your crystal ball for understanding the nature of the roots – in our case, a negative discriminant ( $64 - 100 = -36$ ) told us we were heading for complex conjugate solutions. Dealing with the imaginary unit ' $i$ ' where $i = oxed{\sqrt{-1}}$ is just another tool in our mathematical toolbox. The final simplification to $rac{4 eq 3i}{5}$ and then matching it to the specific $r, s, t$ integer format ( $r=4, s=3, t=5$ ) is crucial for meeting the problem's requirements. This whole process underscores the power and elegance of algebraic methods. Whether you're dealing with real-world problems that can be modeled by quadratic equations or just flexing your math muscles, understanding and applying the quadratic formula is an essential skill. Keep practicing, keep experimenting with different equations, and you'll become a quadratic equation whiz in no time. Don't shy away from those complex solutions; they're just another part of the beautiful mathematical landscape! Stay curious and keep solving!