Solving Simultaneous Equations And Analyzing Roots

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Hey guys! Today, we're diving into some math problems that might seem a bit tricky at first, but I promise, we'll break them down step by step. We're going to tackle simultaneous equations and explore the nature of roots in a quadratic equation. So, grab your thinking caps, and let's get started!

2.1 Solving Simultaneous Equations: $5y - x = 2$ and $x^2 - 3xy + 4y = 4$

Let's dive into solving these simultaneous equations. In this section, we will deal with solving for $x$ and $y$ in the given system of equations. This involves a combination of algebraic techniques to isolate variables and find their values. We'll start with a system of equations:

5y−x=25y - x = 2

x2−3xy+4y=4x^2 - 3xy + 4y = 4

To solve these equations simultaneously, we'll use a mix of substitution and algebraic manipulation. Our main goal here is to find the values of $x$ and $y$ that satisfy both equations. This might sound intimidating, but don't worry, we'll take it one step at a time!

Step 1: Express One Variable in Terms of the Other

The first equation, $5y - x = 2$, looks simpler, so let's use that to express $x$ in terms of $y$. We can rearrange this equation to isolate $x$:

x=5y−2x = 5y - 2

This is a crucial step because now we have $x$ defined in terms of $y$, which we can substitute into the second equation. By isolating $x$, we make it easier to substitute its value into the second equation, simplifying the problem significantly.

Step 2: Substitute into the Second Equation

Now, we'll substitute this expression for $x$ into the second equation, $x^2 - 3xy + 4y = 4$. Replacing $x$ with $(5y - 2)$, we get:

(5y−2)2−3(5y−2)y+4y=4(5y - 2)^2 - 3(5y - 2)y + 4y = 4

This step is where the magic happens! By substituting, we've transformed the second equation into one that only involves $y$. This is a big win because now we can focus on solving for a single variable. It might look a bit messy right now, but trust me, we're on the right track.

Step 3: Expand and Simplify

Next, we need to expand and simplify the equation. Let's start by expanding the squared term and the product:

(25y2−20y+4)−(15y2−6y)+4y=4(25y^2 - 20y + 4) - (15y^2 - 6y) + 4y = 4

Now, let's combine like terms:

25y2−20y+4−15y2+6y+4y=425y^2 - 20y + 4 - 15y^2 + 6y + 4y = 4

10y2−10y+4=410y^2 - 10y + 4 = 4

Simplifying further, we subtract 4 from both sides:

10y2−10y=010y^2 - 10y = 0

This simplified quadratic equation is much easier to handle. We've gone from a complex equation to a manageable one by carefully expanding and combining terms. This is a common technique in algebra, and it's super useful for solving problems like this.

Step 4: Solve for $y$

Now that we have a simplified quadratic equation, $10y^2 - 10y = 0$, we can solve for $y$. First, let's factor out a common factor of $10y$:

10y(y−1)=010y(y - 1) = 0

This gives us two possible solutions for $y$:

10y=0extory−1=010y = 0 ext{ or } y - 1 = 0

So,

y=0extory=1y = 0 ext{ or } y = 1

We've found two possible values for $y$! This is great progress. Each of these values will give us a corresponding value for $x$, which we'll find in the next step. Remember, the key to solving equations is to break them down into smaller, manageable parts.

Step 5: Substitute $y$ Values to Find $x$

Now that we have the values for $y$, we can substitute them back into the equation $x = 5y - 2$ to find the corresponding values for $x$.

For $y = 0$:

x=5(0)−2=−2x = 5(0) - 2 = -2

So, one solution is $(-2, 0)$.

For $y = 1$:

x=5(1)−2=3x = 5(1) - 2 = 3

So, another solution is $(3, 1)$.

We've found two pairs of solutions for $x$ and $y$! This is the final step in solving the simultaneous equations. Always remember to substitute your values back into the original equations to check your work. It's a good habit that helps prevent errors.

Step 6: Verify the Solutions

Finally, let's verify our solutions by substituting them back into the original equations:

For $(-2, 0)$:

5(0)−(−2)=2ext(True)5(0) - (-2) = 2 ext{ (True)}

(−2)2−3(−2)(0)+4(0)=4ext(True)(-2)^2 - 3(-2)(0) + 4(0) = 4 ext{ (True)}

For $(3, 1)$:

5(1)−3=2ext(True)5(1) - 3 = 2 ext{ (True)}

(3)2−3(3)(1)+4(1)=9−9+4=4ext(True)(3)^2 - 3(3)(1) + 4(1) = 9 - 9 + 4 = 4 ext{ (True)}

Both solutions satisfy the original equations, so we're good to go! This step is super important to make sure we haven't made any mistakes along the way. It's like the final checkmark on a job well done.

Final Answer for 2.1

The solutions for the simultaneous equations are $(-2, 0)$ and $(3, 1)$. We did it! Solving simultaneous equations can be challenging, but with a systematic approach, we can tackle any problem. Remember, break it down, step by step, and always check your work!

2.2 Discussing the Nature of Roots of the Equation: $2(x - 3)^2 + 2 = 0$

Now, let's switch gears and discuss the nature of roots for the equation $2(x - 3)^2 + 2 = 0$. In this section, we will delve into determining the nature of the roots of the given quadratic equation. This involves analyzing the discriminant to ascertain whether the roots are real, distinct, real and equal, or non-real (complex). Understanding the nature of roots is crucial in analyzing quadratic equations. It tells us a lot about the solutions and the behavior of the equation.

Step 1: Rewrite the Equation in Standard Form

First, let's rewrite the equation in the standard quadratic form, which is $ax^2 + bx + c = 0$. Expanding the given equation, $2(x - 3)^2 + 2 = 0$, we get:

2(x2−6x+9)+2=02(x^2 - 6x + 9) + 2 = 0

2x2−12x+18+2=02x^2 - 12x + 18 + 2 = 0

2x2−12x+20=02x^2 - 12x + 20 = 0

Now, we have the equation in standard form. This is a critical step because the standard form allows us to easily identify the coefficients $a$, $b$, and $c$, which we'll need to determine the nature of the roots.

Step 2: Simplify the Equation (Optional)

We can simplify the equation by dividing all terms by 2:

x2−6x+10=0x^2 - 6x + 10 = 0

Simplifying the equation makes the coefficients smaller and easier to work with. This step is optional but often helpful in reducing the chances of making errors in subsequent calculations. It's all about making the problem as manageable as possible.

Step 3: Calculate the Discriminant

The discriminant, denoted by $Δ$, is a key factor in determining the nature of roots. The discriminant is given by the formula:

Δ=b2−4acΔ = b^2 - 4ac

For our equation, $x^2 - 6x + 10 = 0$, we have $a = 1$, $b = -6$, and $c = 10$. Let's calculate the discriminant:

Δ=(−6)2−4(1)(10)Δ = (-6)^2 - 4(1)(10)

Δ=36−40Δ = 36 - 40

Δ=−4Δ = -4

The discriminant is negative! This is a crucial piece of information. The value of the discriminant tells us a lot about the nature of the roots, whether they are real, distinct, equal, or complex. Let's see what a negative discriminant means.

Step 4: Interpret the Discriminant

Now that we have the discriminant, $Δ = -4$, we can interpret its value to determine the nature of the roots.

  • If $Δ > 0$, the equation has two distinct real roots.
  • If $Δ = 0$, the equation has two equal real roots (a repeated root).
  • If $Δ < 0$, the equation has two non-real (complex) roots.

Since our discriminant is $Δ = -4$, which is less than 0, the equation has two non-real (complex) roots. This means the solutions involve imaginary numbers. Understanding these conditions is essential for quickly determining the type of solutions we can expect.

Step 5: State the Nature of Roots

Based on our calculation, the discriminant is negative, so the equation $2(x - 3)^2 + 2 = 0$ has two non-real (complex) roots. We've successfully determined the nature of the roots by calculating and interpreting the discriminant. This is a fundamental concept in algebra, and it's super useful for understanding quadratic equations.

Final Answer for 2.2

The equation $2(x - 3)^2 + 2 = 0$ has two non-real (complex) roots. We've covered a lot in this section, from rewriting the equation to calculating the discriminant and interpreting its value. Remember, the discriminant is your best friend when it comes to understanding the nature of roots!

Conclusion

So, guys, we've tackled some serious math problems today! We solved simultaneous equations and discussed the nature of roots. Remember, the key to solving complex problems is to break them down into smaller, manageable steps. Keep practicing, and you'll become math whizzes in no time! If you have any questions, drop them in the comments below. Keep learning and keep shining!