SLC-S22W4//Linear and Quadratic equations equations."
Hello everyone I hope you are all doing good and enjoying life Today I am going to share my assignment on Linear and quadratic Equations I learned a lot from this lecture and I am excited to share it with you all along with my completed assignment task.
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let me do the diffrent between in the linear and Quadratic equation. i am a tutuion teacher so i am explaning these two equation to my students in very simple terms. i am explanaing here i hope you huys will also learn from it.
Linear Equations
A linear equation are very simple. When we graph it it make straight line. The highest power of the variable (like (x) or (y)) is always 1.
General Form of a Linear Equation:
The general form look like this
ax + by + c = 0
Here (a) (b) and (c) are number and (x) and (y) are the variable.
Example of a Linear System:
- (2x + y = 5)
- (x - 3y = 4)
When we solve this we finding values for (x) and (y) that work in both equation.
Quadratic Equations
A quadratic equation is little harder. When we graph it it makes a U shaped curve called a parabola. The highest power of the Variable is 2.
General Form of Quadratic Equation
The general form look like this
ax^2 + bx+c = 0
Here (a) (b) & (c) are numbers and (x) is the Variable.
Example of Quadratic System
- ( x^2 + 3x + 2=0)
- (x^2 - 4x + 4=0)
to solve this we find the value of x that make both equation true.
Main Difference
- Linear equations are simple and make straight line on a graph.
- Quadratic equations are more complex and make curve called parabolas.
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In this task i am going to share the method of solving the quadratic equations and also i am share the pro and corns of these quadratic equations. every quadratic equations own pro and corns my favourite one factoring
1. Factaring
Factaring is when you break the quadratc equation into two smaller expression. For example:
x^2 + 5x +6 = 0 become (x + 2) (x + 3) = 0
Then you set each part to zero to solve:
x + 2 = 0 or x + 3 = 0 Rightarrow x = -2 -3
Pros:
- Quick and simple if the equation is easy to factor.
- Does not require advanced math tools.
Cons:
- Only works if the quadratic can be factored easily.
- Not suitable for equations with complex numbers or decimals.
2. Using the Quadratic Formula
The quadratic formula is:
x = -b √b^2 - 4ac ÷ 2a
It works for any quadratic equation. For example in the equation x^2 + 5x + 6 = 0 , a = 1 b = 5 c = 6.
Pros:
- Works for all quadratic equations even when they can’t be factored.
- Can handle complex solutions when the discriminant b^2 - 4ac is negative.
Cons:
- Involves more calculations which can be tricky without a calculator.
- Mistakes can happen when handling square roots or fractions.
3. Completing the Square
This method involves rewriting the equation to make one side a perfect square. For example:
x^2 + 6x + 9 = 0 becomes (x + 3)^2 = 0
Then solve for x:
x + 3 = 0 Rightarrow x = -3
Pros:
- Helps understand how quadratics work geometrically.
- Works for all quadratics and leads to the quadratic formula.
Cons:
- More steps so it’s slower than other methods.
- Harder for beginners to grasp.
4. Graphing
You graph the equation and find where it crosses the (x)-axis. For example graphing x^2 - 4 = 0 shows the solutions x = 2 and x = -2.
Pros:
- Visual and easy to understand.
- Can give approximate solutions quickly.
Cons:
- Requires graphing tools (like a graphing calculator or software).
- Less accurate for exact solutions especially with decimals or fractions.
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Solve for linear equation 3x + 2 = 11 and show value of x?
Solve for this quadratic equation x^2 + 2x - 6 = 0.
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Scenario 1
Ali has $15 to spend on snack. He buys pack of chip for $3. I want to figure out how much money he has lefts.
Lets call the amount of money Ali has left (x).
The equation is:
x + 3 = 15
Step 1: Understand the Equation
This equation means that if we add the $3 Ali spent on chips to the money he has left it equals $15.
Step 2: Solve for x
To find x we need to isolate it.
Subtract 3 from both sides of the equation:
x + 3 - 3 = 15 - 3
This simplifies to:
x = 12
Step 3: Check the Answer
To double-check I add (x = 12) back into the original equation:
12 + 3 = 15
It’s correct!
Final Answer
Ali has $12 left after buying the pack of chips.
To find the maximum height reached by the ball we need to determined the maximum value of (h(t)) which happens at the vertex of the quadratic equation Here how I would solve it step by step:
Given Equation
The height of the ball as a function of time is:
h(t) = -5t^2 + 20t
This is quadratic equation and since the coeficient of t^2 is negative -5 the graphs is a downward facing parabola meaning it has maximum point.
Step 1: Find the Time at Maximum Height
The time at which the ball reach its maximums height can be found using the formula for the vertex of parabola
t = -b÷2a
Here a = -5 and b = 20. Substitute these values into the formula
t = -20÷2(-5)
t= -20÷-10
t = 2 seconds
Step 2: Find the Maximum Height
Now that we know the ball reache its maximum height at (t = 2) seconds substitute (t = 2) into the Height equation to find (h):
h(t) = -5t^2 + 20t
h(2) = -5(2)^2 + 20(2)
h(2) = -5(4) + 20(2)
h(2) = -20 + 40
h(2) = 20
Final Answer
The maximum height reached by the ball is 20 meters and it happens at 2 seconds.