You know how to solve quadratic equations using algebra, graphs and the quadratic formula. Sometimes one method of solving is more convenient than another method. Describe how you would solve each equation. Give reasons for you answers.
A) 16x^2 = 1600
B) x^2 - 350 = 0
C) x^2 + 10x = 0
I just want to know what you guys think so I have an idea of maybe how to explain my answer.
How I would solve each?
first: take sqrt of each side
second: factor as difference of two squares
third: factor (x(x+10))=0
Yeah, kinda.
I am asking what is the best way to solve each one by either: algebraically, graphing and using the quadratic formula and why you chose that method.
Sure! I'd be happy to explain how to solve each of these quadratic equations using different methods.
A) 16x^2 = 1600:
To solve this equation using algebra, we can start by dividing both sides of the equation by 16 to isolate x^2. This gives us:
x^2 = 100
Next, we take the square root of both sides to solve for x:
x = ±√100
x = ±10
To solve this equation using a graph, we can plot the quadratic equation y = 16x^2 - 1600 on a coordinate plane. The graph of this equation will be a parabola that opens upward. By analyzing the graph, we can see that the parabola intersects the x-axis at x = -10 and x = 10, which are the solutions to the equation.
Lastly, to solve this equation using the quadratic formula, we can rewrite it in the standard form ax^2 + bx + c = 0 by moving 1600 to the other side of the equation:
16x^2 - 1600 = 0
Now we can apply the quadratic formula, which states that for a quadratic equation in the form ax^2 + bx + c = 0, the solutions are given by:
x = (-b ± √(b^2 - 4ac)) / 2a
For our equation, a = 16, b = 0, and c = -1600. Plugging these values into the formula, we get:
x = (0 ± √(0^2 - 4(16)(-1600))) / (2(16))
Simplifying this expression gives us the same solutions as before: x = ±10.
B) x^2 - 350 = 0:
To solve this equation using algebra, we can add 350 to both sides to isolate x^2:
x^2 = 350
Then, we take the square root of both sides to solve for x:
x = ±√350
When using a graph, we can plot the quadratic equation y = x^2 - 350 and observe where the graph intersects the x-axis, which will give us the solutions to the equation.
Using the quadratic formula, we first rewrite the equation in standard form:
x^2 - 350 = 0
Next, we apply the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
For our equation, a = 1, b = 0, and c = -350. Substituting these values into the formula gives us:
x = (0 ± √(0^2 - 4(1)(-350))) / (2(1))
Simplifying this expression gives us x = ±√350.
C) x^2 + 10x = 0:
To solve this equation using algebra, we can factor out x from the left side of the equation:
x(x + 10) = 0
Now, we have two factors, and the equation will be true if either of them equals zero. Therefore, we have two possible solutions:
x = 0
x + 10 = 0
Which gives us x = -10.
When using a graph, we can plot the quadratic equation y = x^2 + 10x and observe where the graph intersects the x-axis.
Lastly, we can use the quadratic formula to solve this equation. Rewriting it in standard form, we get:
x^2 + 10x = 0
Then, we can apply the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
For our equation, a = 1, b = 10, and c = 0. Substituting these values into the formula gives us:
x = (-10 ± √(10^2 - 4(1)(0))) / (2(1))
Simplifying this expression gives us x = 0 and x = -10, which are the same solutions as before.
In summary, for each of these quadratic equations, the solutions can be found using algebra by isolating x, by analyzing the graph, or by applying the quadratic formula. The choice of method depends on personal preference, the complexity of the equation, and the available resources.